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THE 


ELEMENTS  OF  GEOMETRY. 


BY 


GEORGE    BRUCE    HALSTED, 

A.B.,  A.M.,  AND  EX-FELLOW  OF  PRINCETON  COLLEGE;    PH.D.  AND  EX-FELLOW  OF  JOHNS 

HOPKINS  university;    instructor  in   POST-GRADUATE  MATHEMATICS, 

PRINCETON  college;    PROFESSOR  OF  PURE  AND  APPLIED 

MATHEMATICS,  UNIVERSITY  OF  TEXAS. 


NEW  YORK : 
JOHN    WILEY    &    SONS, 

15     ASTOR     PLACE. 
1885.      . 


Copyright,  1885, 
By  JOHN  WILEY  &  SONS. 

CAJORI 


ELECTROTVPED  AND    PRINTED 
BY  RAND,  AVERY,  AND  COMPANY, 


(by  permission) 

TO 

J.    J.     SYLVESTER, 

A.M.,  CAM.;    F.R.S.,  L.  AND  E.;   CORRESPONDING  MEMBER  INSTITUTE  OF  FRANCE;   MEMBER 

ACADEMY  OF  SCIENCES  IN  BERLIN,  GOTTINGEN,  NAPLES,  MILAN,  ST.  PETERSBURG, 

ETC.;    LL.D.,  UNIV.  OF  DUBLIN,  AND  U.  OF  E.;    D.C.L.,  OXFORD;    HON. 

FELLOW  OF  ST.  JOHN'S  COL.,  CAM.;    SAVILIAN  PROFESSOR 

OF  GEOMETRY  IN  THE  UNIVERSITY  OF  OXFORD; 


In  (Grateful  B£membrance 

OF  BENEFITS   CONFERRED   THROUGHOUT   TWO 
FORMATIVE   YEARS. 


918182 


PREFACE. 


In  America  the  geometries  most  in  vogue  at  present  are 
vitiated  by  the  immediate  assumption  and  misuse  of  that 
subtile  term  "  direction ; "  and  teachers  who  know  something 
of  the  Non-Euclidian,  or  even  the  modern  synthetic  geom- 
etries, are  seeing  the  evils  of  this  superficial  "  directional " 
method. 

Moreover  the  attempt,  in  these  books,  to  take  away  by 
definition  from  the  familiar  word  "distance"  its  abstract 
character  and  connection  with  length-units,  only  confuses  the 
ordinary  student.  A  reference  to  the  article  Measurement 
in  the  "  Encyclopaedia  Britannica "  will  show  that  around  the 
word  "distance"  centers  the  most  abstruse  advance  in  pure 
science  and  philosophy.  An  elementary  geometry  has  no 
need  of   the  words  direction  and  distance. 

The  present  work,  composed  with  special  reference  to 
use  in  teaching,  yet  strives  to  present  the  Elements  of 
Geometry  in  a  way  so  absolutely  logical  and  compact,  that 
they  may  be  ready  as  rock-foundation  for  more  advanced 
study. 

Besides  the  acquirement  of  facts,  there  properly  belongs 
to  Geometry  an  educational  value  beyond  any  other  element- 


VI  PREFACE. 


ary   subject.      In   it   the    mind   first    finds    logic   a  practical 
instrument   of   real   power. 

The  method  published  in  my  Mensuration  for  the  treat- 
ment of  solid  angles,  with  my  words  steregon  and  sieradian, 
having  been  adopted  by  such  eminent  authorities,  may  I 
venture  to  recommend  the  use  of  the  word  sect  suggested 
in  the  same  volume } 

From  1877  I  regularly  gave  my  classes  the  method  of 
Book  IX.  In  1883  my  pupil,  H.  B.  Fine,  at  my  suggestion, 
wrote  out  a  Syllabus  of  Spherical  Geometry  on  the  lines  of 
my  teaching,  which  I  have  followed  in  Book  IX. 

The  figures,  which  I  think  give  this  geometry  a  special 
advantage,  owe  all  their  beauty  to  my  colleague.  Professor 
A.  V.  Lane,  who  has  given  them  the  benefit  of  his  artistic 
skill  and  mastery  of  graphics. 

The  whole  work  is  greatly  indebted  to  my  pupil  and 
friend,  Dr.  F.  A.  C.  Perrine.  We  have  striven  after  accuracy. 
Any  corrections  or  suggestions  relating  to  the  book  will  be 
thankfully  received. 

GEORGE    BRUCE    HALSTED. 
2004  Matilde  Street, 

Austin,  Texas. 


CONTENTS. 


THE   ELEMENTS   OF  GEOMETRY. 

BOOK    I. 


CHAPTER  I. 


ON  LOGIC. 


I.  Definitions.  —  Statements. 

ARTICLE  PAGE 

I.    A  statement  defined i 

Immediate  inference  defined  ...  i 

Equivalent  statements  defined  .    .  i 

A  statement  implied i 

Statements  reduced  to  two  terms 

and  a  copula 2 

Logically  simple  sentence ....  2 

Logically  composite  sentence    .    .  2 
Conditional  statements  reduce  to 

the  typical  form 2 

9.    Subject  and  predicate 2 

II.  Definitions.  —  Classes. 

10.  Individual  terms  defined   ....  2 

11.  Class  terms  defined 2 

12.  How  a  class  is  defined 3 

13.  Contradictory  of  a  class     ....  3 

14.  Contradictories  include  the  universe,  3 

III.  The  Universe  of  Discourse. 

15.  Universe  of  discourse  defined    .    .  3 

16.  The  classes  x  and  non-x  still  con- 

tradictory      3 

17.  Different  divisions  into  contradic- 

tories    3 


IV.   Contranominal,  Converse,  Inverse, 
Obverse. 

ARTICLE  PAGE 

18.  Contranominal  defined 

19.  Converse  defined 

20.  Inverse  defined 

21.  Obverse  defined 

22.  Two  theorems  prove  four  .... 

23.  Classes  proved  identical    .... 


24. 

25- 
26. 
27. 
28. 
29. 
30- 


31- 
32. 


V.  On  Theorems. 


A  theorem  defined  .... 
Demonstration  defined  .  , 
Corollary  defined ..... 
Hypothesis  and  conclusion  , 
A  geometric  theorem  defined 
Use  of  type  in  this  work  .  , 
Inverses  of  theorem  with  composite 
hypothesis 6 


VI.  On  Proving  Inverses. 

Inverses  often  proved  geometrically,  6 

Rule  of  identity 6 

Rule  of  inversion 7 

vii 


Vlll 


CONTENTS. 


CHAPTER   II. 


THE   PRIMARY   CONCEPTS   OF  GEOMETRY. 


I.  Definitions  of  Geometric  Magnitudes. 

ARTICLE  PAGE 

34.    Geometry  defined 8 

A  solid  defined 8 

A  surface  defined 8 

A  line  defined 8 

A  point  defined 8 

A  magnitude  defined 8 


35- 
36. 

yi- 
38. 

39. 

40-43 

44. 

45- 
46. 

47. 

48. 
49. 

50. 

51- 
52. 

53- 
54. 
55- 
56. 

57- 
58. 

59- 
60. 
61. 

62. 
63. 

64. 
65. 


Point,  line,  surface,  solid,   de- 
fined, using  motion  ....  8 
Why  space  is  called  tri-dimensional,  9 
Straight  line  defined     .....  9 

A  curve  defined 9 

Line    henceforth    means    straight 

line 9 

A  sect  defined 9 

A  plane  defined 9 

Plane  defined,  using  motion  ...  9 

A  figure  defined 9 

A  plane  figure  defined 9 

Circle  and  center  defined  ....  10 

Radius  defined 10 

Diameter  defined 10 

Arc  defined 10 

Parallel  lines  defined 10 

Angle,  vertex,  arms,  defined  ...  10 

Explement  defined 11 

Equal  angles  defined 11 

Adjacent  angles  and  their  sum  de- 
fined    II 

Straight  angle  defined 12 

When  two  adjacent  angles  equal  a 

straight  angle 12 

Supplement  defined 12 

Supplemental  adjacent  angles    .    .  12 


66.  Right  angle  defined 13 

67.  A  perpendicular  defined    ....  13 

68.  Complement  defined 13 

69.  Acute  angle  defined 13 

70.  Obtuse  angle  defined 13 

71.  Perigon  defined 13 

72.  Sum  of  angles  about  a  point  is  a 

perigon 14 

73.  Reflex  angle  defined 14 

74.  Oblique  angles  and  lines  defined    .  14 

75.  Vertical  angles  defined 14 

76.  Bisector  defined 14 

Tj.    A  trace  defined 14 

•jS.    A  triangle  defined 15 

79.  Vertices  defined 15 

80.  Sides  defined 15 

81.  Interior  angle  defined 15 

82.  Exterior  angle  defined 15 

83.  Congruent  defined 15 

84.  Equivalent  defined 15 

II.  Properties  of  Distinct  Things. 

85-93.    The  so-called  axioms  of  arith- 
metic    I5> 16 

III.  Some  Geometrical  Assumptions  about 
Euclid'' s  Space. 

94-99.    The  so-called  geometric  axioms,  16 

IV.   The  Assumed  Constructions. 

100-103.    The  so-called  postulates .    .    .  17 

Table  of  symbols  used 17 


CHAPTER  III. 

PRIMARY   RELATIONS   OF   LINES,   ANGLES,   AND   TRIANGLES, 
I.  Angles  about  a  Point. 


ARTICLE  PAGE 

104-105.  Right  and  straight  angles  .    18, 19 

106-107.    Perigons 19 

1 08-1 10.    Vertical  angles 20,  21 


II.  Angles  about  Two  Points. 

ARTICLE  PAGE 

III.  Transversal  defined    ....    22 

1 1 2-1 13.    Corresponding   and    alternate 

angles 22 


CONTENTS. 


IX 


III.  Triangles. 

ARTICLE  PAGE 

114.    Equilateral  triangle  defined  ...  24 

Isosceles  triangle  defined  ....  24 

Scalene  triangle  defined    ....  24 

Base  and  vertex  defined    ....  24 
Angles  of  triangle  mean  interior 

angles 24 


[15. 
16. 
:i7. 


ARTICLE  PAGE 

119.  Right-angled  triangle  and  hypoth- 

enuse  defined 

120.  Obtuse-angled  triangle  defined 

121.  Acute-angled  triangle  defined 

122.  Equiangular  triangle  defined 
Homologous  defined     .    .    . 


123. 

124-129,     Congruent  triangles 


CHAPTER   IV. 

PROBLEMS. 


ARTICLE  PAGE 

130.  Problem  defined 30 

131.  Solution  defined 30 


ARTICLE 

132-40.    Deduced  constructions 


PAGE 

30-35 


CHAPTER  V. 

JNEQUALITIES. 


ARTICLE  PAGE 

141.  Sign  of  inequality  explained  .    36 

142-150.    Theorems  on  one  triangle  .   36-39 


ARTICLE  PAGE 

151.  Oblique  sect  defined  ....    39 

152-164.    Obliques  and  triangles    .    .   40-47 


CHAPTER  VI. 

PARALLELS. 


ARTICLE 
165-172. 


Fundamental  theorems 


PAGE     ARTICLE  PAGE 

48-51  I  173-175.    Application  to  the  triangle .   51,52 


CHAPTER  VII. 

TRIANGLES. 


PAGE 


176-179.    Congruent  triangles  .    .    .   53-55 

180.  Conditions  of    congruence  of 

triangles 55 

181.  Locus  defined 56 


ARTICLE  PAGE 

182.  How  to  prove  a  locus     ...    56 

183-186.    Loci 57,58 

187-191.  Intersection  of  loci    .    .    .    58-60 


CHAPTER  VIII. 

POLYGONS. 


I.  Definitions. 

ARTICLE  PAGE 

192.  A  polygon  defined 61 

193.  Its  vertices  defined 61 


ARTICLE 

194.  Its  sides  defined  .... 

195.  Perimeter  defined     .    .    . 

196.  Interior  angles  of  a  polygon 


PAGE 

.  61 
.  61 
.    62 


CONTENTS. 


ARTICLE  PAGE 

197.    Convex  defined 62 

198. 
199. 
200. 
201. 


Diagonals  defined 62 

An  equilateral  polygon  defined  .    .  62 
An  equiangular  polygon  defined    .  62 
A  regular  polygon  defined     ...  63 
Mutually  equilateral  defined ...  63 
Mutually  equiangular  defined    .    .  63 
204,  205.     Polygons  may  be  either  ...  63 
206.    Trigon,  tetragon  or  quadrilateral, 
pentagon,    hexagon,    heptagon, 
octagon,  nonagon,  decagon,  do- 
decagon, quindecagon   ....  64 
The  surface  of  a  polygon  defined  .  64 

Parallelogram  defined 64 

Trapezoid  defined 64 


203. 


207. 
208. 


209. 

II.  General  Properties. 
210-215.    Congruence,  angles    . 


216-220. 

221. 

222. 

223, 

224, 225. 

226. 


III.  Parallelograms. 

Sides  and  angles  . 
A  rectangle  defined 
A  rhombus  defined 
A  square  defined  . 
Properties     .    .    . 


64-66 


67-69 

•  70 

•  70 
.     70 

70,71 


Triangle  altitudes  concurrent .     72 


227-2: 

232. 

233. 

234. 

235- 

236. 


237. 

238-240. 
241-244. 
245-255. 
256,  257. 


258-264. 


PAGE 

Intercepts 73>  74 

Medial  defined 74 

Concurrent  defined     ....  75 

Collinear  defined 75 

Medials  concurrent     •     •     .     .  75 
Centroid,  orthocenter,  circum- 

center,  defined 75 

IV.  Equivalence. 

Equivalence,  how  proved    .     .     76 

Base  and  altitude 76 

Squares 77-79 

Rectangles 80-86 

Complements  of  parallelo- 
grams    87 

V.  Problems. 

Parallelograms    equivalent    to 
polygons 88-90 


VI.  Axial  and  Central  Symmetry. 

265.  Corresponding  points  defined .     91 

266.  Axial  symmetry  defined      .     .     91 

267.  Central  symmetry  defined  .     .     92 

268.  269.     Symmetry  in  one  figiire  .    .   92,  93 


BOOK    II 


RECTANGLES. 


270.  A  continuous  aggregate  defined     .  95 

271.  A  discrete  aggregate  defined ...  95 
272-279.    Counting  and  number     .    .    95, 96 

280.  The    commutative    law    for    addi- 

tion       96 

281.  The  sum  of  two  sects  defined    .    .  97 

282.  The   commutative   law   holds   for 

sects 97 

283.  The   sum   of   two   rectangles  de- 

fined     q8 


ARTICLE  PAGE 

284.  The   commutative   law   holds   for 

rectangles 98 

285.  The  associative  law 99 

286.  The  associative  law  holds  for  sects,  99 

287.  The  associative  law  holds  for  rect- 

angles       99 

288.  The  commutative  law  for  multipli- 

cation  100 

289.  The  commutative  law  for  rectangle 

of  sects 100 


CONTENTS, 


XI 


ARTICLE 

290. 

291-293. 

294,  295. 

296-302. 


PAGE 

The  distributive  law  .  .  .100 
The  distributive  law  holds  for 

sects  and  rectangles  .  100-102 
The  square  on  the  sum  of  two 

sects 102, 103 

Rectangles  and  squares  .    103-106 


ARTICLE  PAGE 

303.  The  projection  of  a  point  de- 

fined      106 

304.  The  projection  of  a  sect  de- 

fined      107 

305-311.    Squares  on  sides    .    ,    .    107-111 
312-314.    Problems 111-113 


BOOK    III 


THE   CIRCLE. 


ARTICLE 

315-319. 

320,321. 

322,  323. 

324- 

325- 

326. 

327. 

328. 

329- 
330-332. 

333»  334. 
335»  336. 
337-340. 
341-343. 
344-348. 
349-353. 
354-356. 

357. 
358,  359. 


360. 

361,  362. 
363- 
364. 
365. 


I.  Primary  Properties. 

PAGE 

Center,  radius,  surface  of  circle,  115 
Point  and  circle     .    . 


16, 


17 


A  secant  

A  chord  defined     .    .    . 
A  segment  defined      .    . 
Expleraental  arcs  defined 
Semicircle  defined  ..... 
Major  arc  and  minor  arc  de- 
fined      

Major  segment  and  minor  seg- 
ment defined 117 

Circle  and  center 118 

Diameter 118 

Congruent  circles 119 

Concentric  circles 120 

Symmetry  in  the  circle  .     120, 121 

Chords 122-124 

Sects 125, 126 

Circumscribed    circle  and  in- 
scribed polygon 126 

Coney clic  defined 126 

Perpendiculars  on  chords    .    .127 


II.  Angles  at  the  Center. 

Angles  subtended  by  arcs  .    .128 

Sector 128 

Angle  subtended  by  a  sect  .    .128 

Inscribed  angle 128 

Corresponding    arcs,    angles, 
sectors,  chords 129 


PAGE 

129, 130 
130-132 
132, 133 


ARTICLE 

366-368.  Sum  of  two  minor  arcs  . 
369-374.  Unequal  arcs  .... 
375-378.    Inscribed  angles     .    ,    . 

III.  Angles  in  Segments. 


379-386.    Angles  made  by  chords  and 

secants 134-138 

IV.  Tangents. 

387,  388.    Tangent  and  point  of  contact 

defined 138 

389-402.    Properties  of  tangents   .     139-143 

V.  Two  Circles. 

403,  404.    Common  chord 144 

405.  Tangent  circles  defined  .    .    .  144 

406.  Common  axis  of  symmetry      .  144 
407-411.     Relations  between  the  center- 
sect,  radii,  and  relative  posi- 
tion      145-147 

VI.  Problems. 

412.  To  bisect  a  given  arc.    .    .    .  148 

413.  Circumscribed  polygon  and  in- 

scribed circle 148 

414.  Escribed  circle  defined    .    .    .148 
415-417.    Problems 149, 150 


Xll 


CONTENTS. 


BOOK    IV 


REGULAR  POLYGONS. 


I.  Partition  of  a  Perigon. 

ARTICLE  PAGE 

418,419.    Successive  bisection  ....  151 

420-422.    Trisection 152 

423-426.    Five  and  fifteen     .    .    .    153-154 

II.   Regular  Polygons  and  Circles. 

427-435.  To  inscribe  and  circum- 
scribe         154-159 

436.  Radius  and  apothem  de- 
fined      159 


ARTICLE  PAGE 

437.  Side  of    inscribed  hexagon 

equals  radius 159 

III.  Least  Perimeter  in  Equivalent  Figures. 
—  Greatest  Surface  in  Isoperimetric  Fig- 
ures. 

438.  Isoperimetric  defined.    .    .    .159 
439-443.     Equivalent   and  isoperimetric 

triangles 160-162 

444-446.    The  circle 162-164 

447-450.    Regular  polygons  .    .    .    165-167 


BOOK   V. 


RATIO  AND    PROPORTION. 


ARTICLE 
451-453- 

454- 
455- 
456. 

457. 

458-460. 
461, 462. 


463- 


Multiples 169,170 

Commensurable  defined .  .  . 
Incommensurable  defined  .  . 
Commensurability  exceptional. 
Greatest     common    submulti- 

ple 

Incommensurable  magnitudes. 
The  side  and  diagonal  of   a 
square    are    incommensur- 
able  172,173 

Scales  of  multiples  defined 


PAGE 

ARTICLE 

?,  170 

464-472. 

.    170 

473-476. 

.    170 

477,478 

,    170 

479- 

.    170 

480-482 

,    171 

483. 

484-491. 

2,173 

492-496. 

.  ^n 

PAGE 

Multiples 173-175 

Scale  of  relation  .  .  .  175, 176 
Ratio,  antecedent,  consequent,  176 
Proportion,  extremes,  means, 

fourth  proportional    .     .     .177 
Equality  of  ratios  defined  .     .177 
Third  proportional,  mean  pro- 
portional    177 

Ratios  compared    .    .    .     178, 179 
Ways  of  changing  a  propor- 
tion       180-182 


BOOK    VI 


RATIO  APPLIED. 


I.  Fundamental  Geometric  Proportions. 

ARTICLE  PAGE 

497-500.    Intercepts  by  parallels    .     183,  184 
501-503.    Segments  of  a  sect     .    .    184-186 


ARTICLE  PAGE 

504,  505.    Rectangles  are  as  their  bases, 

j86,  187 
506.  Angles  are  as  arcs 187 


CONTENTS. 


Xlll 


II.  Similar  Figures. 

PAGE 

Similar  figures  defined    .    .    .188 
Similar  triangles    .    .    .     188-192 
Homologous,  ratio   of   simili- 
tude, similarly  placed      .     .192 
515,516.    Similar  polygons  similarly 

placed 192, 193 

517,518.    Center  of  similitude  ....  193 
519-522.    The  right  triangle     ....  194 


507. 

508-513. 

514. 


ARTICLE  PAGE 

523,  524.    The  bisector  of  an  angle,  195, 196 
525-528.     Harmonic  division 196 

III.  Rectangles  and  Polygons, 
529-539.     Rectangle  of  extremes  equiv- 
alent to  rectangle  of 

means 197-204 

540-547.    Composition  of  ratios      .    204-208 
548-552.    Theorems  and  problems,    208-212 


GEOMETRY  OF  THREE   DIMENSIONS. 

BOOK   VII. 


OF  PLANES  AND   LINES. 


ARTICLE  PAGE 

553.  The  plane  defined 213 

554-556.  Assumptions  about  the  plane  .213 

557-562.  Determination  of  a  plane,  214-216 

563-565.  Intersection  of  planes     .    216,217 

566.  Principle  of  duality    .    ..     .    .217 

567-578.  Perpendicular  to  plane  .    218-224 


ARTICLE 

PAGE 

579-581. 

Projection  on  plane    .    . 

224, 225 

582. 

Inclination  defined     .    . 

.    .  226 

583-593- 

Parallel  planes  .... 

226-231 

594,  595- 

Smallest  sect  between  lines 

,232,233 

596-604. 

Loci  and  angles  made 

by 

planes 

233-238 

BOOK   VIII 


TRI-DIMENSIONAL  SPHERICS. 


ARTICLE 

605-609. 

610. 

611,612. 

613,614. 

615,616. 

617,618. 

619. 

620-624. 

625. 


PAGE 

Sphere,  center,  radius,  defined,  239 
Opposite  points  and  diameter 

defined 240 

Point  and  sphere 240 

Plane  and  sphere  .    .    .    240, 241 
Great  circle  and  small  circle    .  241 

Poles  and  axis 241 

Hemispheres 241 

Great    circles   and   small    cir- 
cles   242,243 

Zone  and  its  bases  defined  .    .  243 


ARTICLE  PAGE 

626-630.    Tangency 243,244 

631,  632.    Intersecting  spheres  ....  245 
633-639.    Sphere  and  tetrahedron  .    246-248 
640, 641.    Globe   and  globe-segment  de- 
fined      248 

642, 643.    Central    and    planar    sym- 

metry 248, 249 

644.            Globe  equivalent   to    tetrahe- 
dron       250 

645,646.    Chords  and  arcs 251 

647-649.    Quadrant 252 


XIV 


CONTENTS. 


650-652.    Spherical  angles  and  perpen- 
dicular arcs   ....    252, 253 


ARTICLE 
653,  654. 


Smallest  line  between  two 
points 254,255 


BOOK   IX. 


TWO-DIMENSIONAL  SPHERICS. 


ARTICLE  PAGE 

655.  Introduction 257 

656.  Two-dimensional  definition  of 

the  sphere 258 

657-662.    Fundamental  properties,  258,259 

663-666.    Definitions 259 

667-675.    Assumptions  with    corolla- 
ries      259-261 

676.  All  spherical  lines  intersect    .  261 

677, 678.    A  lune  and  its  angle  ....  262 

679-684.    Definitions 263, 264 

685, 686.    Perimeter  of  polygon      .    264,  265 
687.            Symmetry   without   congru- 
ence       265 

688-691.    Congruent    spherical    tri- 
angles      266-268 

692-697.    Problems 268-270 


ARTICLE  PAGE 

698-703.    The  line  and  its  pole .    .    270, 271 
704.  Bi-rectangular  triangle    .    .     .  272 

705-707.    Polar  triangles  ....    272,  273 
708-719.    Spherical  triangles     .    .    274-280 

720, 721.    Loci 280, 281 

722,  723.    Symmetrical  triangles  equiva- 
lent   281, 282 

725.  Spherical  excess  defined     .    .  282 

724-728.    Triangles     equivalent     to 

lunes 282,283 

729-731.     Polygons     equivalent     to 

lunes 283, 284 

732.  Equivalent  triangles  on  same 

base 284 

IZZ^TA'     Tangency 285 

735-739.    Theorems  on  circles  .    .    286, 287 


BOOK   X. 


POLYHEDRONS. 


ARTICLE  PAGE 

740-744.    Polyhedron,  faces,  edges,  sum- 
mits, section ....    289,  290 
745-747-    Pyramid,  apex,   base,   lateral 

faces  and  edges      ....  290 

748.  Parallel  polygons 290 

749-757.     Prism,   altitude,   right   prism, 

oblique 290, 291 

758-760.     Parallelopiped,  quader,  cube  .  291 
761.  All  summits  joined  by  one  line,  292 


ARTICLE  PAGE 

762-764.  Descartes'  theorem    ....  293 

765.  Euler's  theorem 294 

766-768.  Ratio  between  quaders   .    295-297 

769.  Parallelopiped   equivalent    to 

quader 298 

770,  771.    Triangular  prism 299 

n^^  773-    Pyramid  sections 300 

774.  Equivalent  tetrahedra    .     .     .  301 

775.  Triangular  pyramid  and  prism,  302 


CONTENTS. 


XV 


BOOK    XL 

MENSURATION,   OR  METRICAL  GEOMETRY. 


CHAPTER   I. 

THE  METRIC   SYSTEM.  —  LENGTH,   AREA. 


ARnCLK  PAGE 

776-778.    Measurement 303 

779-782.  Meter  and  metric  system,  303, 304 

783,784.    Length 304>305 

785-787.    Area 305 


ARTICLE  PAGE 

788.  Area  of  rectangle 305 

789,  790.    Area  of  square 307 

791,  792.  Metric  units  of  surface  .    .    .  308 

793.  Given  hypothenuse  and  side  .  308 


CHAPTER   II. 

RATIO   OF  ANY   CIRCLE  TO   ITS   DIAMETER. 


ARTICLE  PAGE 

794.  To  compute  perimeters    ....  310 

795.  Length  of  circle 312 

796.  Variable  defined 313 

797.  Limit  defined 313 


ARTICLE  PAGB 

798.  Principle  of  limits      .    .    .    .313 

799-802.    Value  of  ir 315?  316 

803,804.    Circular  measure  of  an 

angle 316,317 


CHAPTER  III. 

MEASUREMENT   OF   SURFACES. 


ARTICLE  PAGE 

805, 806.  Area  of  parallelogram  .  .  .318 
807, 808.  Area  of  triangle  .  .  .  318, 319 
809,  810.    Area  of  regular  polygon  and 

circle 320 

811-813.    Area  of  sector  and  sector  of 

annulus 321,322 

814.  Lateral  area  of  prism      .    .    .  322 


815-822.  Cylinder 323-325 

823.  Mantel  defined 325 

824.  Conical  surface 326 

825-832.  Cone  and  frustum .    .    .    326-329 

Z-},1.  Area  of  sphere 330 

834, 835.  Zone 331 


CHAPTER   IV. 

SPACE    ANGLES. 


ARTICLE  PAGE 

^Z^i^yi'    Plane  and  space  angles  de- 
fined      332 

838.  Symmetrical  space  angles      .    .    .  332 

839.  Steregon 332 

840.  Steradian 333 

841.  Space  angle  made  by  two  planes    .  333 

842.  Spherical  pyramid   ......  333 


ARTICLE  PAGE 

843.  Space  angles  proportional     .    .    .  333 

844.  To  transform  any  polyhedral  angle,  334 

845.  846.    Area  of  a  lune 334 

847.  Plane  angle  as  measure  of  dihedral,  335 

848.  Properties  of  space  angles     .    .    .335 
849-851.    Area  of  spherical  triangle  or 

polygon 336 


XVI 


CONTENTS. 


ARTICLE 
852-854. 
855,856. 
857. 
858,  859. 

860-862. 
863-865. 

866-868. 


CHAPTER  V. 


THE  MEASUREMENT  OF  VOLUMES. 


Volume  and  metric  units    .    .337 
Volume  of  quader  and  cube    .  338 
Volume  of  parallelopiped    .    .  338 
Volume  of   prism  or  cyl- 
inder      338,339 

Volume  of  pyramid  or  cone,  339-341 
Prismatoid  defined  .  .  341, 342 
Prismoid 343 


ARTICLE  PAGE 

869,870.    Tetrahedron   and  wedge  are 

prismatoids 343 

871,  872.    Altitude  and  cross-section  of  a 

prismatoid 343 

873,  Volume  of  any  prismatoid  .    .  344 

874,  875.    Frustum  of  pyramid  or  cone  .  346 

876.  Volume  of  globe 347 

Z^^,  878.    similar  solids 347 


Direction 347 

Principle  of  Duality 349 

Linkage 350 

Cross-ratio 352 

Exercises 357-366 


THE  ELEMENTS  OF  GEOMETRY. 


BOOK    I. 


CHAPTER  I. 

ON   LOGIC. 

I.  Definitions.  —  Statements. 

1.  A  statement  is  any  combination  of  words  that  is  either 
true  or  false ;  e.g.  {exempli  gratia^  "  for  example  "), 

All  x^s  are  fs, 

2.  To  pass  from  one  statement  to  another,  with  a  conscious- 
ness that  belief  in  the  first  warrants  belief  in  the  second,  is  to 
infer. 

3.  Two  statements  are  equivalent  when  one  asserts  just  as 
much  as  the  other,  neither  more  nor  less ;  e.g.,  A  equals  By  and 
B  equals  A. 

4.  A  statement  is  implied  in  a  previous  statement  when  its 
truth  follows  of  necessity  from  the  truth  of  the  previous  state^ 
ment;  e.g.. 

All  x's  are  ys  implies  some  fs  are  x*s. 


THE  ELEMENTS  OF  GEOMETRY. 


5.  Any  declarative  sentence  can  be  reduced  to  one  or  more 
simple  statements,  each  consisting  of  three  parts,  namely, 
two  terms  or  classes,  and  a  copula,  or  relation,  connecting 
them. 

A  equals  B,  x  is  7,  are  simple  examples  of  this  typical  form 
of  all  statements.  Here  x  and  y  each  stand  for  any  word  or 
group  of  words  that  may  have  the  force  of  a  substantive  in 
naming  a  class  ;  e.g.,  x\^y  may  mean  all  the  x^^  are  y^  ;  man 
is  mortal,  means,  all  men  are  mortals. 

6.  A  sentence  containing  only  one  such  statement  is  a 
logically  simple  sentence ;  e.g.,  Man  is  the  only  picture-making 
animal. 

7.  A  sentence  that  contains  more  than  one  such  statement 
is  a  logically  composite  sentence ;  e.g.,  A  and  B  are  respectively 
equal  to  C  and  D,  is  composite,  containing  the  two  statements 
A  equals  C,  and  B  equals  D. 

8.  Statements  that  are  expressly  conditional,  such  as,  if  A 
is  B,  then  C  is  D,  reduce  to  the  typical  form  as  soon  as  we  see 
that  they  mean 

{The  consequence  of  M)  is  N. 

9.  In  the  typical  form,  bird  is  biped,  x  is  y,  we  call  x  and  y 
the  terms  of  the  statement;  the  first  being  called  the  subject, 
and  the  last  the  predicate. 

II.  Definitions. — Classes. 

10.  Terms  that  name  a  single  object  are  called  individual 
terms ;  e.g.,  Newton. 

11.  Terms  that  name  any  one  of  a  group  of  objects  are 
called  class  terms ;  e.g.,  mafti  crystal.  The  name  stands  for 
any  object  that  has  certain  properties,  and  these  properties  are 
possessed  in  common  by  the  whole  group  for  which  the  name 
stands. 


ON  LOGIC, 


12.  A  class  is  defined  by  stating  enough  properties  to  decide 
whether  a  thing  belongs  to  it  or  not. 

Thus,  "rational  animal"  was  given  as  a  definition  of  "man." 

13.  If  we  denote  by  x  the  class  possessing  any  given  prop- 
erty, all  things  not  possessing  this  property  form  another  class, 
which  is  called  the  contradictory  of  the  first,  and  is  denoted  by 
non-;ir,  meaning  "  not  x ; "  e.g.,  the  contradictory  of  animate  is 
inanimate. 

14.  Any  one  thing  belongs  either  to  the  class  x  or  to  the 
class  non-;r,  but  no  thing  belongs  to  both. 

It  follows  that  X  is  just  as  much  the  contradictory  of  nonvr 
as  non-;r  is  of  x.  So  any  class  y  and  the  class  non-7  ^^^  mutu- 
ally the  contradictories  of  each  other,  and  both  together  include 
all  things  in  the  universe ;  e.g.,  unconscious  and  conscious. 

III.  The  Universe  of  Discourse. 

15.  In  most  investigations,  we  are  not  really  considering  all 
things  in  the  world,  but  only  the  collection  of  all  objects  which 
are  contemplated  as  objects  about  which  assertion  or  denial 
may  take  place  in  the  particular  discourse.  This  collection  we 
call  our  universe  of  discourse,  leaving  out  of  consideration,  for 
the  time,  every  thing  not  belonging  to  it. 

Thus,  in  talking  of  geometry,  our  terms  have  no  reference 
to  perfumes. 

16.  Within  the  universe  of  discourse,  whether  large  or 
small,  the  classes  x  and  non-;ir  are  still  mutually  contradictory, 
and  every  thing  is  in  one  or  the  other ;  e.g.,  within  the  universe 
mammals,  every  thing  is  man  or  brute. 

17.  The  exhaustive  division  into  x  and  non-;ir  is  applicable 
to  any  universe,  and  so  is  of  particular  importance  in  logic. 
But  a  special  universe  of  discourse  may  be  capable  of  some 
entirely  different  division  into  contradictories,  equally  exhaus- 
tive.    Thus,  with  reference  to  any  particular  magnitude,  all 


THE  ELEMENTS  OF  GEOMETRY. 


magnitudes  of  that  kind  may  be  exhaustively  divided  into  the 
contradictories, 

Greater  than,  equal  to,  less  than. 

IV.  Contranominal,  Converse,  Inverse,  Obverse. 

i8.  If  X  and  y  are  classes,  our  typical  statement  x  \s  y  means, 
if  a  thing  belongs  to  class  x,  then  it  also  belongs  to  class  y\ 
e.g.,  Man  is  mortal,  means,  to  be  in  the  class  men,  is  to  be  in 
the  class  mortals. 

If  the  typical  statement  is  true,  then  every  individual  x 
belongs  to  the  class  y :  hence  no  x  belongs  to  the  class  non-y, 
or  no  thing  not  y  is,  3.  thing  x ;  that  is,  every  non-y  is  nonvt; : 
e.g.,  the  immortals  are  not-human. 

The  statements  x  is  y,  and  non-^  is  non-;ir,  are  called  each 
the  contranominal  form  of  the  other. 

Though  both  forms  express  the  same  fact,  it  is,  nevertheless, 
often  of  importance  to  consider  both.  One  form  may  more 
naturally  connect  the  fact  with  others  already  in  our  mind,  and 
so  show  us  an  unexpected  depth  and  importance  of  meaning. 

19.  Since  x  is  y  means  all  the  x's  are  j/'s,  the  class  y  thus 
contains  all  the  individuals  of  the  class  x,  and  may  contain 
others,  besides.  Some  of  the  ys,  then,  must  be  ^s.  Thus, 
from  **a  crystal  is  solid  "  we  infer  "some  solids  are  crystals." 

This  guarded  statement,  some  y  is  Xy  is  called  the  logical 
converse  of  x  is  y.     It  is  of  no  importance  in  geometry. 

20.  If,  in  the  true  statement  x  is  y,  we  simply  interchange 
the  subject  and  predicate,  without  any  restriction,  we  get  the 
inverse  statement  y  is  x,  which  may  be  false. 

In  geometry  it  often  happens  that  inverses  are  true  and 
important.  When  the  inverse  is  not  true,  this  arises  from  the 
circumstance  that  the  subject  of  the  direct  statement  has  been 
more  closely  limited  than  was  requisite  for  the  truth  of  the 
statement. 


ON  LOGIC. 


21.  The  contranominal  of  the  inverse,  namely,  non-;tr  is 
non-^,  is  called  the  obverse  of  the  original  proposition. 

Of  course,  if  the  inverse  is  true,  the  obverse  is  true,  and 
vice  versa.  To  prove  the  obverse,  amounts  to  the  same  thing 
as  proving  the  inverse.  They  are  the  same  statement,  but  may 
put  the  meaning  expressed,  in  a  different  light  to  our  minds. 

22.  If  the  original  statement  is  x  is  j/,  its  contranominal  is 
noii-y  is  non-Xy  its  inverse  is  y  is  Xy  its  obverse  is  non-x  is  non-y. 
The  first  two  are  equivalent,  and  the  last  two  are  equivalent. 

Thus,  of  four  such  associated  theorems  it  will  never  be 
necessary  to  demonstrate  more  than  two,  care  being  taken  that 
the  two  selected  are  not  contranominal. 

23.  From  the  truth  of  either  of  two  inverses,  that  of  the 
other  cannot  be  inferred.  If,  however,  we  can  prove  them 
both,  then  the  classes  x  and  y  are  identical. 

A  perfect  definition  is  always  invertable. 

V.  On  Theorems. 

24.  A  theorem  is  a  statement  usually  capable  of  being 
inferred  from  other  statements  previously  accepted  as  true. 

25.  The  process  by  which  we  show  that  it  may  be  so  in- 
ferred is  called  the  proof  or  the  demonstration  of  the  theorem. 

26.  A  corollary  to  a  theorem  is  a  statement  whose  truth 
follows  at  once  from  that  of  the  theorem,  or  from  what  has 
been  given  in  the  demonstration  of  the  theorem. 

27.  A  theorem  consists  of  two  parts,  —  the  hypothesis,  or 
that  which  is  assumed,  and  the  conclusion,  or  that  which  is 
asserted  to  follow  therefrom. 

28.  A  geometric  theorem  usually  relates  to  some  figure, 
and  says  that  a  figure  which  has  a  certain  property  has  of 
necessity  also  another  property ;  or,  stating  it  in  our  typical 
form,  X  is  j,  "a  figure  which  has  a  certain  property"  is  "a 
figure  having  another  specified  property.'* 


THE  ELEMENTS  OF  GEOMETRY. 


The  first  part  names  or  defines  the  figure  to  which  the 
theorem  relates ;  the  last  part  contains  an  additional  property. 
The  theorem  is  first  stated  in  general  terms,  but  in  the  proof  we 
usually  help  the  mind  by  a  particular  figure  actually  drawn  on 
the  page ;  so  that,  before  beginning  the  demonstration,  the  the- 
orem is  restated  with  special  reference  to  the  figure  to  be  used. 

29.  Type.  —  Beginners  in  geometry  sometimes  find  it  diffi- 
cult to  distinguish  clearly  between  what  is  assumed  and  what 
has  to  be  proved  in  a  theorem. 

It  has  been  found  to  help  them  here,  if  the  special  enuncia- 
tion of  what  is  given  is  printed  in  one  kind  of  type;  the 
special  statement  of  what  is  required,  in  another  sort  of  type ; 
and  the  demonstration,  in  still  another.  In  the  course  of  the 
proof,  the  reason  for  any  step  may  be  indicated  in  smaller  type 
between  that  step  and  the  next. 

30.  When  the  hypothesis  of  a  theorem  is  composite,  that  is, 
consists  of  several  distinct  hypotheses,  every  theorem  formed 
by  interchanging  the  conclusion  and  one  of  the  hypotheses  is 
an  inverse  of  the  original  theorem. 

VI.   On  Proving  Inverses. 

31.  Often  in  geometry  when  the  inverse,  or  its  equivalent, 
the  obverse,  of  a  theorem,  is  true,  it  has  to  be  proved  geomet- 
rically quite  apart  from  the  original  theorem.  But  if  we  have 
proved  that  every  x  is  y,  and  also  that  there  is  but  one  individ- 
ual in  the  class  y^  then  we  infer  that  y  is  x.  The  extra-logical 
proof  required  to  establish  an  inverse  is  here  contained  in  the 
proof  that  there  is  but  one  y. 

Rule  of  Identity. 

32.  If  it  has  been  proved  that  x  is  j,  that  no  two  ;tr's  are  the 
same  /,  and  that  there  are  as  many  individuals  in  class  x  as  in 
class  /,  then  we  infer  y  is  x. 


ON  LOGIC, 


Rule  of  Inversion. 

33.  If  the  hypotheses  of  a  group  of  demonstrated  theorems 
exhaustively  divide  the  universe  of  discourse  into  contradic- 
tories, so  that  one  must  be  true,  though  we  do  not  know  which, 
and  the  conclusions  are  also  contradictories,  then  the  inverse 
of  every  theorem  of  the  group  will  necessarily  be  true. 

Examples  occur  in  geometry  of  the  following  type :  — 

If  a  is  greater  than  by  then  c  is  greater  than  d. 

If  a  is  equal  to  by  then  c  is  equal  to  d. 

If  a  is  less  than  by  then  c  is  less  than  d. 

Three  such  theorems  having  been  demonstrated  geometri- 
cally, the  inverse  of  each  is  always  and  necessarily  true. 

Take,  for  instance,  the  inverse  of  the  first ;  namely,  when 
c  is  greater  than  dy  then  a  is  greater  than  b. 

This  must  be  true ;  for  the  second  and  third  theorems 
imply  that'  if  a  is  not  greater  than  by  then  c  is  not  greater 
than  d. 


THE  ELEMENTS  OF  GEOMETRY. 


CHAPTER  IL 


THE  PRIMARY  CONCEPTS   OF   GEOMETRY. 


I.   Definitions  of  Geometric  Magnitudes. 

34.  Geometry  is  the  science  which  treats  of  the  properties 
of  space. 

35.  A  part  of  space  occupied  by  a  physical  body,  or  marked 
out  in  any  other  way,  is  called  a  Solid. 


3 


D 


36.  The  common  boundary  of  two  parts  of  a  solid,  or  of  a 
solid  and  the  remainder  of  space,  is  a  Surface. 

37.  The  common  boundary  of  two  parts  of  a  surface  is  a 
Line. 

38.  The  common  boundary  of  two  parts  of  a  line  is  a  Point. 

39.  A  Magnitude  is  any  thing  which  can  be  added  to  itself 
so  as  to  double. 

40.  A  point  has  position  without  magnitude. 

41.  A  line  may  be  conceived  of  as  traced  or  generated  by 
a  point  on  a  moving  body.  The  intersection  of  two  lines  is  a 
point. 

42.  A  line  on  a  moving  body  may  generate  a  surface.  The 
intersection  of  two  surfaces  is  a  line. 


THE  PRIMARY  CONCEPTS  OF  GEOMETRY.  9 

43.  A  surface  on  a  moving  body  may  generate  a  solid. 

44.  We  cannot  picture  any  motion  of  a  solid  which  will 
generate  any  thing  else  than  a  solid. 

Thus,  in  our  space  experience,  we  have  three  steps  down 
from  a  solid  to  a  point  which  has  no  magnitude,  or  three  steps 
up  from  a  point  to  a  solid ;  so  our  space  is  said  to  have  three 
dimensions. 

45.  A  Straight  Line  is  a  line  which  pierces  space  evenly,  so 

that  a  piece  of  space  from  along  one  side  of  it   ^ 

will  fit  any  side  of  any  other  portion. 

46.  A  Curve  is  a  line  no  part  of  which  is    ^^ "^>. 

straight.  a  curve. 

47.  Take  notice:  the  word  "line,"  unqualified,  will  hence- 
forth mean  "  straight  line." 

48.  A  Sect  is  the  part  of  a  line  between  two  definite  points. 


49.  A  Plane  Surfacey  or  a  Plane,  is  a  surface  which  divides 
space  evenly,  so  that  a  piece  of  space  from  along  one  side  of  it 
will  fit  either  side  of  any  other  portion. 

50.  A  plane  is  generated  by  the  motion  of  a  line  always 
passing  through  a  fixed  point  and  leaning  on  a  fixed  line. 


51.  A  Figure  is  any  definite  combination  of  points,  lines, 
curves,  or  surfaces. 

52.  A  Plane  Figure  is  in  one  plane. 


10 


THE  ELEMENTS  OF  GEOMETRY, 


53.  If  a  sect  turns  about  one  of  its  end  points,  the  other 
end  point  describes  a  curve  which  is  called  a  Circle,  The 
fixed  end  point  is  called  the  Center  of  the  Circle, 

c 


54.  The  Radius  of  a  Circle  is  a  sect  drawn  from  the  center 
to  the  circle. 

55.  A  Diameter  of  a  Circle  is  a  sect  drawn  through  the 
center,  and  terminated  both  ways  by  the  circle. 

56.  An  Arc  is  a.  part  of  a  circle. 


57.  Parallel  Lines  are  such  as  are  in  the  same  plane,  and 
which,  being  produced  ever  so  far  both  ways,  do  not  meet. 


a  J) 

58.  When  two  lines  are  drawn  from  the  same  point,  they 
are  said  to  contain,  or  to  make  with  each  other,  an  Angle. 


THE  PRIMARY  CONCEPTS  OF  GEOMETRY.  II 

The  point  is  called  the  Vertex,  and  the  lines  are  called  the 
Arms,  of  the  angle.  A  line  drawn  from  the  vertex,  and  turning 
about  the  vertex  in  the  plane  of  the  angle  from  the  position  of 
coincidence  with  one  arm  to  that  of  coincidence  with  the  other, 
is  said  to  turn  through  the  angle ;  and  the  angle  is  greater  as 
the  quantity  of  turning  is  greater. 


59.  Since  the  line  can  turn  from  the  one  position  to  the 
other  in  either  of  two  ways,  two  angles  are  formed  by  two 
lines  drawn  from  a  point. 

Each  of  these  angles  is  called  the  Explement  of  the  other. 
If  we  say  two  lines  going  out  from  a  point  form  an  angle,  we 


\^ 


are  fixing  the  attention  upon  one  of  the  two  explemental  angles 
which  they  really  form  ;  and  usually  we  mean  the  smaller  angle. 

60.  Two  angles  are  called  Equal  if  they  can  be  placed  so 
that  their  arms  coincide,  and  that  both  are  described  simul- 
taneously by  the  turning  of  the  same  line  about  their  common 
vertex. 

61.  If  two  angles  have  the  vertex  and  an  arm  in  common, 
and  do  not  overlap,  they  are  called  Adjacent  Angles ;  and  the 
angle  made  by  the  other  two  arms  on  the  side  toward  the  com- 
mon arm  is  called  the  Sum  of  the  Adjacent  Angles,     Thus, 


12 


THE  ELEMENTS  OF  GEOMETRY. 


using  the  sign   "4-  for  the  word  "angle,"  the  sign  of  equality 
(=),  and  the  sign  of  addition  (+,  plus), 


%.  AOC  =  ^  AOB  +  t  BOC. 

62.  A  Straight  Angle  has  its  arms  in  the  same  line,  and  on 
different  sides  of  the  vertex. 


63.  The  sum  of  two  adjacent  angles  which  have  their 
exterior  arms  in  the  same  line  on  different  sides  of  the  vertex 
is  a  straight  angle. 


64.  When  the  sum  of  any  two  angles  is  a  straight  angle, 
each  is  said  to  be  the  Supplement  of  the  other. 


O  O2  Oi  it 

65.    If  two  supplemental  angles  be  added,  their  exterior 


THE  PRIMARY  CONCEPTS  OF  GEOMETRY. 


13 


arms  will  form  one  line ;  and  then  the  two  angles  are  called 
Supplemental  Adjacent  Angles, 


66.  A  Right  Angle  is  half  a  straight  angle. 

67.  A  Perpendicular  X,o  a  line  is  a  line  that  makes  a  right 
angle  with  it. 


68.  When  the  sum  of  two  angles  is  a  right  angle,  each  is 
said  to  be  the  Complement  of  the  other. 


69.  An  Acute  Angle  is  one  which  is  less  than  a  right  angle. 

70.  An  Obtuse  Angle  is  one  which  is  greater  than  a  right 
angle,  but  less  than  a  straight  angle. 

71.  The  whole  angle  which  a  sect  must  turn  through,  about 
one  of  its  end  points,  to  take  it  all  around  into  its  first  position, 


14 


THE  ELEMENTS  OF  GEOMETRY. 


or,  in  one  plane,  the  whole  amount  of  angle  round  a  point,  is 
called  a  Perigon. 


J  ^ 


72.  Since  the  angular  magnitude  about  a  point  is  neither 
increased  nor  diminished  by  the  number  of 
lines  which  radiate  from  the  point,  the  sum 
of  all  the  angles  about  a  point  in  a  plane 
is  a  perigon. 

73.  A  Reflex  Angle  is  one  which  is  greater  than  a  straight 
angle,  but  less  than  a  perigon. 

74.  Acute,  obtuse,  and  reflex  angles,  in  distinction  from 
right  angles,  straight  angles,  and  perigons,  are  called  Oblique 
Angles ;  and  intersecting  lines  which  are  not  perpendicular  to 
each  other  are  called  Oblique  Lines. 

75.  When  two  lines  intersect,  a  pair  of  angles  contained  by 
the  same  two  lines  on  different  sides  of  the  vertex,  having  no 
arm  in  common,  are  called  Vertical  Angles. 


76.  That  which  divides  a  magnitude  into  two  equal  parts  is 
said  to  halve  or  bisect  the  magnitude,  and  is  called  a  Bisector. 

77.  If  we  imagine  a  figure  moved,  we  may  also  suppose  it 
to  leave  its  outline,  or  Trace^  in  the  first  position. 


THE  PRIMARY  CONCEPTS  OF  GEOMETRY. 


IS 


78.  A  Triangle  is  a  figure  formed  by  three  lines,  each  inter- 
secting the  other  two. 


79.  The  three  points  of  intersection  are  the  three  Vertices 
of  the  triangle. 

80.  The  three  sects  joining  the  vertices  are  the  Sides  of  the 
triangle.     The  side  opposite  A  is  named  a ;  the  side  opposite 

81.  An  Interior  Angle  oi  a  triangle  is  one  between  two  of 
the  sects. 


82.  An  Exterior  Angle  of  a  triangle  is  one  between  either 
sect  and  a  line  which  is  a  continuation  of  another  side. 

83.  Magnitudes  which  are  identical  in  every  respect  except 
the  place  in  space  where  they  may  be,  are  called  Congruent. 

84.  Two  magnitudes  are  Equivalettt  which  can  be  cut  into 
parts  congruent  in  pairs. 

II.  Properties  of  Distinct  Things. 

85.  The  whole  is  greater  than  its  part. 

86.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

87.  Things  which  are  equal  to  the  same  thing  are  equal  to 
one  another. 


1 6  THE  ELEMENTS  OF  GEOMETRY, 

88.  If  equals  be  added  to  equals,  the  sums  are  equal. 

89.  If  equals  be  taken  from  equals,  the  remainders  are 
equal. 

90.  If  to  equals  unequals  are  added,  the  sums  are  unequal, 
and  the  greater  sum  comes  from  adding  the  greater  magnitude. 

91.  If  equals  are  taken  from  unequals,  the  remainders  are 
unequal,  and  the  greater  remainder  is  obtained  from  the  greater 
magnitude. 

92.  Things  that  are  double  of  the  same  thing,  or  of  equal 
things,  are  equal  to  one  another. 

93.  Things  which  are  halves  of  the  same  thing,  or  of  equal 
things,  are  equal  to  one  another. 


III.  Some  Geometrical  Assumptions  about  Euclid's  Space. 

94.  A  solid  or  a  figure  may  be  imagined  to  move  about  in 
space  without  any  other  change.  Magnitudes  which  will  coin- 
cide with  one  another  after  any  motion  in  space,  are  congruent ; 
and  congruent  magnitudes  can,  after  proper  turning,  be  made 
to  coincide,  point  for  point,  by  superposition. 

95.  Two  lines  cannot  meet  twice ;  that  is,  if  two  lines  have 
two  points  in  common,  the  two  sects  between  those  points 
coincide. 

96.  If  two  lines  have  a  common  sect,  they  coincide  through- 
out. Therefore  through  two  points,  only  one  distinct  line  can 
pass. 

97.  If  two  points  of  a  line  are  in  a  plane,  the  line  lies  wholly 
in  the  plane. 

98.  All  straight  angles  are  equal  to  one  another. 

99.  Two  lines  which  intersect  one  another  cannot  both  be 
parallel  to  the  same  line. 


TABLE   OF  SYMBOLS. 


17 


IV.   The  Assumed  Constructions. 

100.  Let  it  be  granted  that  a  line  may  be  drawn  from  any- 
one point  to  any  other  point. 

loi.  Let  it  be  granted  that  a  sect  or  a  terminated  line  may 
be  produced  indefinitely  in  a  line. 

102.  Let  it  be  granted  that  a  circle  may  be  described  around 
any  point  as  a  center,  with  a  radius  equal  to  a  given  sect. 

103.  Remark.  —  Here  we  are  allowed  the  use  of  a  straight 
edge,  not  marked  with  divisions,  and  the  use  of  a  pair  of  com- 
passes ;  the  edge  being  used  for  drawing  and  producing  lines, 
the  compasses  for  describing  circles  and  for  the  transferrence 
of  sects. 

But  it  is  more  important  to  note  the  implied  restriction, 
namely,  that  no  construction  is  allowable  in  elementary  geom- 
etry which  cannot  be  effected  by  combinations  of  these  primary 
constructions. 


SYMBOLS    USED. 


~  similar. 

=  equivalent. 

^  congruent. 

II  parallel  to. 

_L  perpendicular  to. 

-i-  plus. 

.'.  therefore. 

^  angle. 


<  less  than. 

>  greater  than-. 

A  triangle. 

£y  parallelogram* 

O  circle, 

rt.  right. 

St.  straight. 


i8 


THE  ELEMENTS  OF  GEOMETRY. 


CHAPTER  III. 

PRIMARY    RELATIONS    OF    LINES,    ANGLES,    AND    TRIANGLES. 

I.   Angles  about  a  Point. 

Theorem  I. 

104.  All  right  angles  are  equal. 

Such  a  proposition  is  proved  in  geometry  by  showing  it  to 
be  true  of  any  two  right  angles  we  choose  to  take. 

^  3 


G 


J£        D  C      O 


The  hypothesis  will  be,  that  we  have  any  two  right  angles  ; 
as,  for  instance,  right  4.  A  OB  and  right  i.  FHG. 

By  the  definition  of  a  right  angle  (^6)^  the  hypothesis  will 
mean,  :i  A  OB  is  half  a  straight  angle  at  O,  and  ^  FHG  is  half 
a  straight  angle  at  H. 

The  conclusion  is,  that  ^  AOB  and  ^  FHG  are  equal. 

The  proof  consists  in  stating  the  equality  of  the  two  straight 
angles  of  which  the  angles  AOB  and  FHG  are  halves,  referring 
to  the  assumption  (in  98)  that  all  straight  angles  are  equal, 
and  then  stating  that  therefore  the  right  ^  AOB  equals  the 
right  4  FHG,  because,  by  93,  things  which  are  halves  of  equal 
things  are  equal. 


RELATIONS  OF  LINES,   ANGLES,   AND    TRIANGLES.         1 9 

Now,  we  may  restate  and  condense  this  as  follows,  using 
the  abbreviations  rt.  for  "  right,"  and  st.  for  "  straight,"  and  the 
symbol  .'.  for  the  word  "therefore"  : 

Hypothesis.     -4.  A  OB  is  rt  ^  ;  also  ^  FUG  is  rt  ^, 
Conclusion.    ^  AOB  =  ^  FHG. 

Proof.     ^  AOB  \s  half  a  st.  ^  ;  also  ^  FBG  is  half  a  st.  ^. 
(66.  A  right  angle  is  half  a  straight  angle.) 

By  98,  all  straight  angles  are  equal, 

.-.     ^AOB  =  ^FHG. 
(93.  Halves  of  equals  are  equal.) 

105.  Corollary.  From  a  point  on  a  line,  there  cannot  be 
more  than  one  perpendicular  to  that  line. 

Theorem  II. 

106.  All  perigons  are  equal. 


Xb^ 


JC  i<  ^ 


Hypothesis.     4-  ^  ^^  '-^  ^  perigon  ;  also  ^  DHD  is  a  perigon. 
Conclusion.     4A0A=  4.  DHD. 

Proof.    Any  line  through  the  vertex  of  a  perigon  divides  it  into  two 
straight  angles. 

By  98,  all  straight  angles  are  equal, 

.*.     Perigon  at  O  equals  perigon  at  H. 
{92.  Doubles  of  equals  are  equal.) 

107.   Corollary.     From  the  preceding  demonstration,  it 
follows  that  half  a  perigon  is  a  straight  angle. 


20  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  III. 

1 08.  If  two  lines  intersect  each  other ^  the  vertical  angles  are 
equal. 


Hypothesis.    AC  and  BD  are  two  lines  intersecting  at  O. 

Conclusions.     ^  AOB  =  ^  COD. 
^BOC  ^  y^DOA. 
Proof.     Because  the  sum  of  the  angles  AOB  and  BOC  is  the 
angle  AOC,  and  by  hypothesis  AO  and  DC  are  in  one  line, 
.-.     ^  AOB  +  4.  BOC  =  St.  4. 
In  the  same  way, 

^BOC  +  4COD  =  St.  2^, 
.-.     i.AOB^-4  BOC  =  i.  BOC  +  4.  COD. 

Take  away  from  these  equal  sums  the  common  angle  BOC^  and  we 

have 

4.  AOB  =  4COD. 

(89.  If  equals  be  taken  from  equals,  the  remainders  are  equal.) 
In  the  same  way  we  may  prove 

4  BOC  =  4DOA. 


Exercises,  i.  Two  angles  are  formed  at  a  point  on  one 
side  of  a  line.  Show  that  the  lines  which  bisect  these  angles 
contain  a  right  angle. 


RELATIONS  OF  LINES,   ANGLES,   AND    TRIANGLES.         21 


Theorem  IV. 

109.  If  four  lines  go  out  from  a  point  so  as  to  make  each 
angle  equal  to  the  one  not  adjacent  to  it,  the  four  lines  will  form 
only  two  intersecting  li7ies. 


Hypothesis.  Let  OA,  OB,  OC,  OD,  be  four  lines,  with  the 
common  point  O ;  and  let  4.  AOB  =  ^  COD,  and  4.BOC  = 
^  DOA. 

Conclusions.    AO  and  OC  are  in  one  line. 
BO  and  OD  are  in  one  line. 

Proof.     ^  AOB  +  4.  BOC  -f  4  COD  +  4.  DOA  =  a  perigon, 

(71.  The  sum  of  all  the  angles  about  a  point  in  a  plane  is  a  perigon.) 

By  hypothesis,  4  AOB  =  4  COD,  and  4  BOC  =  4  DOA, 

.*.    twice  4  AOB  ■+■  twice  4  BOC  =  a  perigon; 

.-.     4  AOB  +  4  BOC  =  a  st.  4. 
(107.  Half  a  perigon  is  a  straight  angle.) 

.'.  AOand  OC  form  one  line. 
(65.  If  two  supplemental  angles  be  added,  their  exterior  arms  will  form  one  line.) 

In  the  same  way  we  may  prove  that  4  AOB  and  4  DOA  are  sup- 
plemental adjacent  angles,  and  so  that  BO  and  OD  form  one  Hne. 

no.  Corollary.  The  four  bisectors  of  the  four  angles 
formed  by  two  lines  intersecting,  form  a  pair  of  lines  perpen- 
dicular to  each  other. 


22  THE  ELEMENTS  OF  GEOMETRY. 


II.  Angles  about  Two  Points. 

111.  A  line  cutting  across  other  lines  is  called  a  Transversal. 

112.  If  in  a  plane  two  lines  are  cut  in  two  distinct  points 
by  a  transversal,  at  each  of  the  points  four  angles  are  deter- 
mined. 


Of  these  eight  angles,  four  are  between  the  two  given 
lines  (namely,  ^,  J,  «,  ^),  and  are  called  Interior  Angles ;  the 
other  four  lie  outside  the  two  lines,  and  are  called  Exterior 
Angles. 

Angles,  one  at  each  point,  which  lie  on  the  same  side  of  the 
transversal,  the  one  exterior,  the  other  interior,  like  /  and  a, 
are  called  Corresponding  Angles. 

Two  angles  on  opposite  sides  of  the  transversal,  and  both 
interior  or  both  exterior,  like  J  and  a,  are  called  Alternate 
Angles. 


Theorem  V. 

113.  If  two  corresponding  or  two  alternate  angles  are  equal y 
or  if  two  interior  or  two  exterior  angles  on  the  same  side  of  the 
transversal  are  supplemental,  then  every  angle  is  equal  to  its,  cor- 
respondijtg  and  to  its  alternate  angle,  aiid  is  supplemental  to  the 
angle  on  the  same  side  of  the  transversal  which  is  interior  or 
exterior  according  as  the  first  is  interior  or  exterior. 


RELATIONS  OF  LINES,   ANGLES^  AND    TRIANGLES.        23 

First  Case.  —  Hypothesis,    -^.a  —  4.  i. 
Conclusions.    2<dJ  =  2(^z  =  ^j  =  ^^. 

%.a  -^  4.4  =  4.1  ■\-  4.d  ^  4.2  +  4.C  =  4d  -\-  4s  =  St.  4, 


c    d 


Proof.    4a  =  4c,   4h  ^  4d,   41  =  43,    42  =  44. 

{108.  If  two  lines  intersect,  the  vertical  angles  are  equal.) 
Moreover,  since,  by  hypothesis,  4^  —  4-  ^i  ^^^ir  supplements  are 
equal,  ox  4  ^  —  4-  ^^ 

(98.  All  straight  angles  are  equal.) 
(8g.  If  equals  be  taken  from  equals,  the  remainders  are  equal.) 

/.     4a  =  41  =  4c  =  43,     4b  =  42  =  4d=4  4, 
and  St.  ^  =  4  ^  +  4  d  =  4  a  -\-  4  4  =  4  i  -\-  4  d  =  4  2  +  4  c 

(88.  If  equals  be  added  to  equals,  the  sums  are  equal.) 

Second  Case.  If,  instead  of  tsvo  corresponding,  we  have  given  two 
alternate  angles  equal,  we  substitute  for  one  its  vertical,  which  gives  the 
First  Case  again. 

Third  Case.  —  Hypothesis.    4  ^  +  4  4  =  ^^-  4* 
But  ^  ^  +  2^  7  =  St.  ^ , 

.-.     4a  =  41, 
which  gives  again  the  First  Case. 

Fourth  Case.  —  Hypothesis.    2J^  7  +  2j^  ^  =  ^f.  ^ . 
But^  «  +  ^  ^=  St.  4, 

.-.     4a  =  41, 
which  gives  again  the  First  Case. 


24 


THE  ELEMENTS  OF  GEOMETRY. 


III.  Triangles. 

114.  An  Equilateral  Triangle  is  one  in  which  the  three  sides 
are  equal. 


115.  An  Isosceles  Triangle  is  one  which  has  two  sides 
equal. 

116.  A  Scalene  Triangle  has  no  two 
sides  equal. 

117.  When  one  side  of  a  triangle  has  to  be  distinguished 
from  the  other  two,  it  may  be  called  the  Base ;  then  that  one 
of  the  vertices  opposite  the  base  is  called  the  Vertex. 

118.  When  we  speak  of  the  angles  of  a  triangle,  we  mean 
the  three  interior  angles. 

119.  A  Right-angled  Triangle  has  one  of  its  angles  a  right 
angle.  The  side  opposite  the  right  angle  is  called  the  Hypothe- 
nuse. 


120.  An  Obtuse-angled  Triangle  has  one  of  its  angles  obtuse. 

121.  An  Acute-angled  Triangle  has  all  three  angles  acute. 


122.  An  Equiangular  Triangle  is  one  which  has  all  three 
angles  equal. 

123.  When  two  triangles  have  three  angles  of  the  one 
equal  respectively  to  the  three  angles  of  the  other,  a  pair  of  equal 
angles  are  called  Homologous  Angles.  The  pair  of  sides  oppo- 
site homologous  angles  are  called  Homologous  Sides. 


RELATIONS  OF  LINES,  ANGLES,   AND    TRIANGLES. 


25 


Theorem  VI. 

124.  Two  triangles  are  coftgruent  if  tivo  sides  and  the  in- 
eluded  angle  in  the  one  ai'e  equal  respectively  to  two  sides  ajzd 
the  included  angle  in  the  other. 

M 


I  3L 

Hypothesis.    ABC  and  LMN  two  triangles,  with  AB  —  LM, 

BC  =  MN,     4.B  =  ^M. 

Conclusion.  The  two  triangles  are  congruent;  or,  using  A  for 
"triangle,"  and  =  for  "congruent," 

t.ABC^  t.LMN, 

Proof.  Apply  the  triangle  LMN  to  A  ABC  in  such  a  manner  that 
the  vertex  M  shall  rest  on  B,  and  the  side  ML  on  BA,  and  the  point 
N  on  the  same  side  of  BA  as  C. 

Then,  because  the  side  ML  equals  BA,  the  point  L  will  rest  upon 
A ;  because  ^  M  =  :^  B,  the  side  MJV  will  rest  upon  the  line  BC; 
because  the  side  MN  equals  BC,  the  point  JV  will  rest  upon  the 
point  C.  Now,  since  the  point  L  rests  upon  A,  and  the  point  IV  rests 
upon  C,  therefore  the  side  LN  coincides  with  the  side  A  C. 

(95.  If  two  lines  have  two  points  in  common,  the  two  sects  between  those  points 

coincide.) 

Therefore  every  part  of  one  triangle  will  coincide  with  the  corre- 
sponding part  of  the  other,  and  the  two  are  congruent. 

(94.  Magnitudes  which  will  coincide  are  congruent.) 

125.  In  any  pair  of  congruent  triangles,  the  homologous 
sides  are  equal. 


26  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  VII. 

126.  In  an  isosceles  triangle  the  angles  opposite  the  equal 
sides  are  equal. 


Hypothesis.    ABC  a  triangle,  with  AB  =  BC. 

Conclusion.     4.  A  =  :^C. 

Proof.  Imagine  the  triangle  ABC  to  be  taken  up,  turned  over,  and 
put  down  in  a  reversed  position ;  and  now  designate  the  angular  points 
A'  (read  A  prime),  B\  C,  to  distinguish  the  triangle  from  its  trace 
ABC  left  behind. 

Then,  in  the  triangles  ABC,  C'B'A, 

AB  =  C'B'    and    BC  =  B'A', 

since,  by  hypothesis,  AB  =  CB. 

Also  ^B  =  ^B', 

.'.     ^A  =  ^  C 
(124.  Triangles  are  congruent  if  two  sides  and  the  included  angle  are  equal  in  each.) 

But  ^  C  =  ^  C, 

.-.     ^A  =  :^C 
{87.  Things  equal  to  the  same  thing  are  equal  to  one  another.) 

127.  Corollary.  Every  equilateral  triangle  is  also  equi- 
angular. 

Exercises.  2.  The  bisectors  of  vertical  angles  are  in  the 
same  line. 


RELATIONS   OF  LINES,   ANGLES,   AND    TRIANGLES.         2/ 


Theorem  VIII. 

128.  Two  triangles  are  congrtient  if  two  angles  and  the  in- 
cluded side  in  the  one  are  equal  respectively  to  two  angles  and 
the  i7icluded  side  iti  the  other. 


Hypothesis.  ABC  and  LMN  two  iriangles,  with  ^  A  =  ^  Z, 
^C  =  ^JV,    AC=LN. 

Conclusion.    The  two  triangles  are  congruent. 

A  ABC  ^  A  LMN. 

Proof.  Apply  the  triangle  LMN  to  the  triangle  ABC  so  that  the 
point  L  shall  rest  upon  A,  and  the  side  LN  lie  along  the  side  A  C,  and 
the  point  M  lie  on  the  same  side  of  AC  as  B. 

Then,  because  the  side  AC  =  ZN, 

.*.     point  iV  will  rest  upon  point  C. 
Because  ^  Z  =  ^  A, 

.'.    side  ZAf  will  rest  upon  the  line  AB, 
Because  ^  JV  =  ^  C, 

/.     side  JVM  will  rest  upon  the  line  CB. 

Because  the  sides  ZM  and  NM  rest  respectively  upon  the  lines  AB 
and  CB, 

.*.     the  vertex  M,  resting  upon  both  the  lines  AB  and  CB,  must  rest 
upon  the  vertex  B,  the  only  point  common  to  the  tvvo  lines. 

Therefore  the  triangles  coincide  in  all  their  parts,  and  are  congruent 


.,-] 


28 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   IX. 

129.   Two  triangles  are  congruent  if  the  three  sides  of  the 
one  are  equal  respectively  to  the  three  sides  of  the  other. 


Hypothesis.     Triangles  ABC   and   LMN,    having  AB  =  ZM, 

BC  =  MJV,  and  CA  =  JVZ. 

Conclusion,     a  ABC  ^  a  ZMJV. 

Proof.  Imagine  A  ABC  to  be  applied  to  A  ZMN  in  such  a  way 
that  ZN coincides  with  A C,  and  the  vertex  M  falls  on  the  side  oi  AC 
opposite  to  the  side  on  which  B  falls,  and  join  BM. 

Case  I.     When  BM  cuts  the  sect  AC. 


Then  in  A  ABM,  because,  by  hypothesis,  AB  =  AM, 

.:     ^  ABM  =  ^  AMB. 
(126.  In  an  isosceles  triangle  the  angles  opposite  the  equal  sides  are  equal.) 

And  for  the  same  reason,  in  A  BCM,  because  BC  =  CM, 
.'.     ^  CBM  =  ^  CMB. 

Therefore  ^  ABM  +  ^  CBM  =  ^  AMB  +  ^  CMB; 
(88.  If  equals  be  added  to  equals,  the  sums  are  equal.) 

that  is,  ^ABC^4.  AMC, 

:.     A  ABC  ^  J^  ZMN. 
{124.  Triangles  are  congruent  if  two  sides  and  the  included  angle  are  equal  in  each. 


RELATIONS  OF  LINES,   ANGLES,  AND    TRIANGLES.        29 

Case  II.     When  BM  passes  through  one  extremity  of  the 
sect  AC,  as  C. 


Then,  by  the  first  step  in  Case  I., 

^B  =  ^M', 
and  for  the  same  reason  as  before, 

A  ABC  ^  A  LMN. 


Case  III.    When  BM  falls  beyond  the  extremity  of  the 
sect  AC. 


Then  in  A  ABMy  because,  by  hypothesis,  AB  =  AM, 
/.     ^  ABM  =  ^  AMB. 
And  in  A  BCM,  because  BC  =  CM, 

/.     ^  CMB  =  ^  CBM, 

Therefore  the  remainders,  ^  ABC  =  ^  AMC; 

(89.  If  equals  be  taken  from  equals,  the  remainders  are  equal.) 

and  so,  as  in  Case  I.,  A  ABC  ^  A  LMN, 


30  THE  ELEMENTS  OF  GEOMETRY. 


CHAPTER  IV. 

PROBLEMS. 

130.  A  Problem  is  a  proposition  in  which  something  is 
required  to  be  done  by  a  process  of  construction,  which  is 
termed  the  Solution. 

131.  The  solution  of  a  problem  in  elementary  geometry 
consists,  — 

(i)  In  indicating  how  the  ruler  and  compasses  are  to  be 
used  in  effecting  the  construction  required. 

(2)  In  proving  that  the  construction  so  given  is  correct. 

(3)  In  discussing  the  limitations,  which  sometimes  exist, 
within  which  alone  the  solution  is  possible. 

Problem  I. 

132.  To  describe  an  equilateral  triangle  upon  a  given  sect. 

,. (T.. . 


^\ 


Given,  the  sect  AB, 

Required,  to  describe  an  equilateral  triangle  on  AB. 


PROBLEMS.  31 


Construction.    With  center  A  and  radius  AB^  describe  the  circle 
BCD. 

With  center  B  and  radius  BA^  describe  O  A  CF 
(using  O  for  "  circle  "). 

(102.  A  circle  may  be  described  with  any  center  and  radius.) 

Join  a  point  C,  at  which  the  circles  cut  one  another,  to  the  points 
^and^. 

(100.  A  line  may  be  drawn  from  any  one  point  to  any  other  point.) 

Then  will  ABC  be  an  equilateral  triangle. 
Proof.     Because  A  is  the  center  of  O  BCD, 

.'.     AB  =  AC,  being  radii  of  the  same  circle  ; 
and  because  B  is  the  center  of  the  O  ACB, 

.-.     BA  =  BC,  being  radii  of  the  same  circle. 
Therefore  AC  =  AB  =  BC, 

(87.  Things  equal  to  the  same  thing  are  equal  to  one  another.) 

and  an  equilateral  triangle  has  been  described  on  AB» 

Problem  II. 
133.  On  a  giveii  line,  to  mark  off  a  sect  equal  to  a  given  sect. 


^  J!>\ 


Given,  AB,  a  lino ;  a,  a  sect 

Required,  to  mark  off  a  sect  on  AB  equal  to  a. 

Construction.  From  any  point  (7  as  a  center,  on  AB,  describe 
the  arc  of  a  circle  with  a  radius  equal  to  a. 

If  D  be  the  point  in  which  the  arc  intersects  AB,  then  CD  will  be 
the  required  sect. 

Proof.  All  the  radii  of  the  circle  around  O  are,  by  construction, 
equal  to  a. 

OD  is  one  of  these  radii,  therefore  it  is  equal  to  a. 


THE  ELEMENTS  OF  GEOMETRY. 


Problem  III. 
134.  To  bisect  a  given  angle. 


Given,  ihe  angle  AOD. 

Required,  to  bisect  it. 

Construction.  From  (9  as  a  center,  with  any  radius,  OA,  describe 
the  arc  of  a  circle,  cutting  the  arms  of  the  angle  in  the  points  A  and  B. 

Join  AB,  On  AB,  on  the  side  remote  from  (9,  describe,  by  132,  an 
equilateral  triangle,  ABC. 

Join  OC.     The  line  OC  will  bisect  the  given  angle  AOD. 

Proof.     In  the  triangles  CAO  and  CBO  we  have 

OA  =  OB, 

CA  =  CB  by  construction, 
and  the  side  CO  common. 

.-.     A  CAO  ^  A  CBO, 

(129.  Triangles  with  the  three  sides  respectively  equal  are  congruent.) 

.-.     ^  COA  =  ^  COB, 

.'.     CO  is  ihe  Usector  of  ^  A  OB. 


Exercises,  i.  Describe  an  isosceles  triangle  having  each 
of  the  equal  sides  double  the  base. 

2.  Having  given  the  hypothenuse  and  one  of  the  sides  of  a 
right-angled  triangle,  construct  the  triangle. 


PROBLEMS,  33 


Problem  IV. 

135.   Through  a  given  point  on  a  given  line,  to  draw  a  perpen- 
dicular to  this  line. 


^\  C  i         :b 


Given,  ihe  line  AB,  and  the  point  C  in  it 
Required,  to  draw  from  C  a  perpendicular  to  AB. 
Solution.    By  1 34,  bisect  the  st.  2^  A  CB, 

Exercises.     5.  Solve  134  without  an  equilateral  triangle. 
6.  Divide  a  given  angle  into  four  equal  parts. 

3.  Find  a  point  in  a  line  at  a  given  distance  from  a  givea 
point.     When  is  the  problem  impossible } 

4.  Find  a  point  in  a  line  equally  distant  from  two  given 
points  without  the  line. 

5.  From  a  given  point  without  a  given  line,  draw  a  line 
making  with  the  given  line  an  angle  equal  to  one-half  a  right 
angle. 

6.  y4  is  a  point  without  a  given  line,  EC.  Construct  an 
isosceles  triangle,  having  A  as  vertex,  whose  base  shall  lie 
along  EC,  and  be  equal  to  a  given  sect. 

II.  If  a  right-angle  triangle  have  one  acute  angle  double 
the  other,  the  hypothenuse  is  double  one  side. 


34 


THE  ELEMENTS  OF  GEOMETRY. 


Problem  V. 
136.   To  bisect  a  given  sect. 


A 

#      I      * 

/         •         *« 

/  I  « 

/  '  • 

*  '  \ 

c : 1 


Given,  the  sect  AB. 

Required,  to  bisect  it. 

Construction.  On  AB  describe  an  equilateral  triangle  ABCj  by 
132. 

By  134,  bisect  the  angle  ACB  by  the  line  meeting  AB  in  D,  Then 
AB  shall  be  bisected  in  Z>. 

Proof.     In  the  triangles  A  CD  and  BCD 

^ACD=  ^BCD, 
and  AC  =  BC  by  construction,  and  CD  is  common. 

.-.      AD  =  BD, 
(124.  Triangles  are  congruent  if  two  sides  and  the  included  angle  are  equal  in  each.) 

AB  is  bisected  at  D. 

137.  Corollary  I.  The  line  drawn  to  bisect  the  angle  at 
the  vertex  of  an  isosceles  triangle,  also  bisects  the  base,  and  is 
perpendicular  to  it. 

138.  Corollary  II.  The  line  drawn  from  the  vertex  of  an 
isosceles  triangle  to  bisect  the  base,  is  perpendicular  to  it,  and 
also  bisects  the  vertical  angle. 

Exercises.  12.  Construct  a  right-angled  isosceles  triangle 
on  a  given  sect  as  hypothenuse. 


PROBLEMS.  35 


Problem  VI. 

i39»  From  a  point  without  a  given  line^  to  drop  a  perpendicu- 
lar upon  the  line. 


?€ 


i^ 


Given,  Me  line  AB,  and  the  point  C  without  it 

Required,  to  drop  from  C  a  perpendicular  upon  AB. 

Construction.  Take  any  point,  Z>,  on  the  other  side  of  AB  from 
C,  and,  by  102,  from  the  center  C,  with  radius  CDy  describe  the  arc 
FDG,  meeting  AB  at  F  and  G. 

By  136,  bisect  FG  at  H.    Join  CH, 

C^  shall  be  ±^^ 

(using  ±  for  the  words  "perpendicular  to "). 

Proof.  Because  in  the  triangles  CHF  and  CHGy  by  construction, 
CF  =  CG,  and  HF  =  HG,  and  CH  is  common, 

.-.     ^  CBF  =  ^  C^6J. 

(129.  Triangles  with  three  sides  respectively  equal  are  congruent.) 

.-.  4.  FHC,  being  half  the  st.  4.  FHG,  is  a  rt.  ^  ;  and  CH  is 
perpendicular  to  AB. 

140.  Corollary.  The  line  drawn  from  the  vertex  of  an 
isosceles  triangle  perpendicular  to  the  base,  bisects  it,  and  also 
bisects  the  vertical  angle. 

Exercises.  13.  Instead  of  bisecting  FGy  would  it  do  to 
bisect  the  angle  FCG } 


36  THE  ELEMENTS  OF  GEOMETRY. 


CHAPTER  V. 

INEQUALITIES. 

141'  The  symbol  >  is  called  the  Sign  of  Inequality y  and 
a>  b  means  that  a  is  greater  than  b\  so  a  <d  means  that  a  is 
less  than  d. 

Theorem  X. 

142.  An  exterior  angle  of  a  triangle  is  greater  than  either  of 
the  two  opposite  interior  angles. 


G 


Hypothesis.    Let  the  side  AB  of  the  triangle  ABC  be  produced 
to  D. 

Conclusions.     %.  CBD  >  ^  BCA. 

4.CBD>  ^BAC. 

Proof.    By  136,  bisect  BC  in  H. 

By  100  and  10 1,  join  AH^  and  produce  it  to  F',  making,  by  133, 
HF  =  AH. 

By  100,  join  BF, 


INEQUALITIES.  37 

Then,  in  the  triangles  AHC  and  FHB,  by  construction, 
CH  =  HB,     AH  =  HF,     and     4.  AHC  =  ^  FHB ; 
(108.  Vertical  angles  are  equal.) 

/.     A  AHC  ^  A  FHB, 
(124.  Triangles  are  congruent  if  two  sides  and  the  included  angle  are  equal  in  each.) 

.-.     4  HCA  =  4  HBF. 
Now,  4.  CBD  >  4  HBF, 

(85.  The  whole  is  greater  than  its  part.) 
.-.     4  CBD  >  4.  HCA, 
Similarly,  if  CB  be  produced  to  G,  it  may  be  shown  that 
4ABG>4  BAC, 
.-.    the  vertical  4  CBD  >  4  BAC. 


Theorem  XL 

143.  Any  two  angles  of  a  triangle  are  together  less  than  a 
straight  angle,  jb 


Hypothesis.    Let  ABC  be  any  A. 

Conclusion.     4  A  4-  ^  ^  <  st.  ^ . 
Proof.     Produce  CA  to  D. 
Then  4  BAD  >4B, 
(142.  An  exterior  angle  of  a  triangle  is  greater  than  either  opposite  interior  angle.) 

.-.     4  BAD  +  4  BAC>4  B  -{-  4  BAC, 
.',    st.4>4B-\-4  A, 

144.  Corollary  I.     No  triangle  can  have  more  than  one 
right  angle  or  obtuse  angle. 

145.  Corollary  II.     There  can  be  only  one  perpendicular 
from  a  point  to  a  line. 


38  THE  ELEMENTS  OF  GEOMETRY, 


Theorem  XII. 

146.  If  one  side  of  a  triangle  be  greater  than  a  second^  the 
angle  opposite  the  first  must  be  greater  than  the  angle  opposite^ 
the  second.  ^ 


Hypothesis,     a  ABC,  with  side  a  >  side  c. 

Conclusion.     ^  BA  C>  4.C. 

Proof.     By  133,  from  a  cut  off  BD  =  c. 

By  100,  join  AD. 

Then,  because  BD  =  c,  :.     ^  BDA  =  ^  BAD. 

(126.  In  an  isosceles  triangle  the  angles  opposite  the  equal  sides  are  equal.) 

And,  by  142,  the  exterior  ^  BDA  of  A  CD  A  >  the  opposite  inte- 
rior ^  C,  .-.    also     ^  BAD  >^C. 
Still  more  \s^BAC>:^C. 

147.  Corollary.  If  one  side  of  a  triangle  be  less  than  the 
second,  the  angle  opposite  the  first  will  be  less  than  the  angle 
opposite  the  second. 

For,  if  ^  <  ^,     :.  b  >  a\     :.  by  146,  4.  B>4.A,     .\   '4.A<:4.B. 

148.  Inverses.     We  have  now  proved, 

By  146,  \i  a>  b,  .'.     4.  A  >  4.  B  \ 

Byi26,  if«  =  ^,  .-.     ^A—^B'j 

By  147,  ii  a  <  b,  .*.     :^  A  <  ^  B. 

Therefore,  by  33,  the  inverses  are  necessarily  true,  namely, 

If  :^A  >  ^B,        .'.     a  >  b; 

If  ^A  =  ^B,         .-.     a  =:  b; 

If  4  A  <  4.B,         .-.     a  <  b. 

149.  Corollary.  An  equiangular  triangle  is  also  equi- 
lateral. 


INEQUALITIES.  39 


Theorem  XIII. 

150.   The  perpendicular  is   the  least  sect  between  a  given 
point  and  a  given  line. 


Ji   ^ 


Hypothesis,     lei  A  he  a  given  point,  and  BC  a  given  line. 

Construction.     By  139,  from  A  drop  AD  A.BC. 
By  100,  join  A  to  Fy  any  point  oi  BC  except  D. 
Conclusion.    AD  <  AF. 
Proof.    Since,  by  construction,  ^  ADF  is  rt.  4. , 

:.     -4.  ADF  >  i.  AFD, 
(143.  Any  two  angles  of  a  triangle  are  together  less  than  a  straight  angle.) 

.-.    AF>AD. 

(148.  If  angle  ADF  is  greater  than  angle  AFD,  therefore  side  AF  is  greater  than 

side  AD.) 

151.  Except  the  perpendicular,  any  sect  from  a  point  to  a 
line  is  called  an  Oblique. 


Exercises.  14.  Determine  the  least  path  from  one  point 
to  another,  subject  to  the  condition  that  it  shall  meet  a  given 
straight  line. 

15.  The  four  sides  of  a  quadrilateral  figure  are  together 
greater  than  its  two  diagonals. 


40 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XIV. 

152.  Two  obliques  from  a  point,  making  equal  sects  from  the 
foot  of  the  i)erpendiculary  are  equal,  and  make  equal  angles  with 
the  line. 


Hypothesis.     AC  X.  BD. 
BC  =  DC. 

Conclusions.    AB  =  AD. 

4.  ABC  =  4.  ADC. 
Proof.     AACB^t:^ACD. 

(124.  Triangles  are  congruent  if  two  sides  and  the  included  angle  are  equal  in  each.) 

153-  Corollary.     For  every  oblique,  there  can  be  drawn 
one  equal,  and  on  the  other  side  of  the  perpendicular. 


Exercises.  16.  ABC  is  a  triangle  whose  angle  A  is 
bisected  by  a  line  meeting  BC  at  D.  Prove  AB  greater  than 
BD,  and  A  C  greater  than  CD. 

17.  Prove  that  the  sum  of  the  sects  from  any  point  to  the 
three  angles  of  a  triangle  is  greater  than  one-half  the  perime- 
ter of  the  triangle. 

18.  The  two  sides  of  a  triangle  are  together  greater  than 
twice  the  line  drawn  from  the  vertical  angle  to  the  middle 
point  of  the  base. 


INEQUALITIES. 


41 


Theorem  XV. 

154.  Of  any  two  obliques  betwee^t  a  given  point  and  line^ 
that  which  makes  the  greater  sect  from  the  foot  of  the  perpen- 
dicular is  the  greater. 


Hypothesis.    AC  1.  BD, 

CB  >  CF, 

Conclusion.    AF  <  AB. 

Proof.    By  153,  take  F  on  the  same  side  of  the  perpendicular  as 
B  'j  then,  in  A  AFCy  because,  by  hypothesis,  ^  ACF  is  rt., 

/.     ^  AFC  <  rt.  ^, 
(143.  Any  two  angles  of  a  triangle  are  together  less  than  a  straight  angle.) 

.-.     ^  AFB  >  rt.  ^, 

since  ^  AFC  +  ^  AFB  =  st.  ^. 

/.     ^  ABF  <  i  AFB, 

(143.  Any  two  angles  of  a  triangle  are  together  less  than  a  straight  angle.) 

/.    AF<AB. 
(148.  If  angle  A  is  less  than  angle  By  therefore  a  is  less  than  b.) 

^55'  Corollary.     No  more  than  two  of  all  the  sects  can 
be  equal. 

Proof.    For  no  two  sects  on  the  same  side  of  the  perpendicular 
can  be  equal. 


42  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XVI. 

156.  Any  two  sides  of  a  triangle  are  together  greater  than 
the  third  side. 


Jr/ 


D"'" 


Hypothesis.    Let  ABC  be  a  triangle. 

Conclusions,    a  +  b>c. 

a  -{-  ob. 

b  -\-  oa. 
Proof.     By  loi,  produce  BC  to  D;  making,  by  133,  CD  =  b. 
By  100,  join  AD. 
Then,  because  AC  =  CD,    ■ 

.'.     t  CDA  =  :^  CAD, 

Now,  ^  BAD  >  4.  CAD, 

(85.  The  whole  is  greater  than  its  part.) 

.*.    also  ^  BAD  >  4.  BDA. 
By  148,  .-.     BD>BA, 

But  BD  =  a  +  b,  and  BA  is  c, 

.*.     a  -{-  b>c. 
Similarly  it  may  be  shown  that  a  -j-  ob,  and  that  b  +  oa, 

157.  Corollary.     Any  side  of  a  triangle  is  greater  than 
the  difference  between  the  other  two  sides. 


INEQUALITIES.  43 


Theorem  XVII. 

158.  If  from  the  ends  of  one  of  the  sides  of  a  triangle  two 
sects  be  drawn  to  a  point  within  the  triangle^  these  will  be 
together  less  than  the  other  two  sides  of  the  triangle^  but  will 
contain  a  greater  angle. 


Hypothesis.    D  is  a  point  within  A  ABC. 
Conclusions.     (I.)  AB  +  BC>AD  +  DC. 

(II.)   4.  ADC      >B. 
Proof.     (I.)  Join  AD  and  CD. 
Produce  CD  to  meet  AB  in  F. 
Then  {CB  +  BF)  >  {CF),  and  \_DF  +  FA;\  >  [DA-]. 

(156.  In  a  triangle  any  two  sides  are  together  greater  than  the  third.) 

/.     CB  -h  BA  =  (CB  +  BF)  ^  FA>  {CF)  +  FA 

=^  CD  +  IDF  +  FA'\  >CD  +  IDA']. 

(II.)  Next,  in  A  ^i^Z? 

:^ADC>:^AFD. 

(142.  Exterior  angle  of  a  triangle  is  greater  than  opposite  interior  angle.) 
And  for  the  same  reason,  in  A  CFB 

4AFD>4.B,        /.     i.ADC>4.B. 


Exercises.     19.  The  perimeter  of  a  triangle  is  greater  than 
the  sum  of  sects  from  a  point  within  to  the  vertices. 


44  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XVIII. 

159.  If  two  triangles  have  two  sides  of  the  one  equal  respec- 
tively to  two  sides  of  the  other y  but  the  included  angle  of  the 
first  greater  than  the  included  angle  of  the  second^  then  the  third 
side  of  the  first  will  be  greater  than  the  third  side  of  the  second. 


Hypothesis.    ABC  and  DEF  are  two  fr/ang/es,  in  which 
AB  =  DE, 
BC  =  EF, 
^  ABC  >  4.  DEF, 

Conclusion.    AC>  DF. 

Proof.  Place  the  triangles  so  that  EF  shall  coincide  with  BC,  and 
the  point  D  fall  on  the  same  side  of  j5C  as  A. 

By  134,  bisect  ^  ABD,  and  let  G  be  the  point  in  which  the  bisector 
meets  A  C. 

By  100,  join  DG. 

Then,  in  the  triangles  ABG  and  BDG,  by  construction,  ^  ABG  = 
-4.  DBG'i  by  hypothesis,  AB  =  BD;  and  BG  is  common. 

.-.     A  ABG  ^  A  DBG, 
(124.  Two  triangles  are  congruent  if  two  sides  and  the  included  angle  are  equal  in 

each.) 

.-.    AG  =  DG, 

.-.     CA  =  CG  i-  GA  =  CG  -{-  GD. 


INEQUALITIES. 


45 


But  CG  -\-  GD>  CD, 

(156.  Any  two  sides  of  a  triangle  are  together  greater  than  the  third.) 

.-.     CA>FD, 


160.  Corollary.     U  c  =  f^  a  =  dy  and  i.B<^  E,  then 
i.E>'4.B\  and  so,  by  1 59, 


e>by        .'.    b<e. 


161.  Inverses.     We  have  now  proved,  when  a  •=.  d^  and 

By  159,  \i4.B  >  4.E, 
By  124,  if  ^^  =  ;^^, 
By  160,  if  ^B  <  4.E, 

Therefore,  by  33,  Rule  of  Inversion, 


3  >  <?j 
b=e; 
b  <  e. 


lib  >  e, 

:.     ^B  >  ^E] 

If  d  =  e, 

.-.     ^B  =  ^E; 

If  b  <  e, 

.'.     ^B  <  ^E, 

Exercises.     20.  In  a  ABC,  AB  <  AC.      D  is  the  middle 
point  of  BC.     Prove  the  angle  ADB  acute. 


46  THE  ELEMENTS  OF  GEOMETRY. 


Problem  VI  I. 

162.  To  construct  a  triangle  of  which  the  sides  shall  be  equal 
to  three  given  sects,  but  any  two  whatever  of  these  sects  must  be 
greater  than  the  third. 


\ 

^                     ' 

\ 

» 

:d\ 

0          \     : 

a 

Z-^" 

3F 

^ » ' 

'••  *m^m' 

Tr 

c    - 

Given,  fhe  three  sects  a,  b,  c,  any  two  whatever  greater  than 
the  third. 

Required,  to  make  a  triangle  having  its  sides  equal  respectively  to  a, 
b,  c. 

Construction.  On  a  line  DF,  by  133,  take  a  sect  OG  equal  to  one 
of  the  given  sects,  as  b.  From  O  as  a,  center,  with  a  radius  equal  to 
one  of  the  remaining  sects,  as  ^,  describe,  by  102,  a  circle. 

From  (9  as  a  center,  with  a  radius  equal  to  the  remaining  sect  a, 
describe  an  arc  intersecting  the  former  circle  at  X.  Join  XO  and  XG, 
KGO  will  be  the  triangle  required. 

Proof.  By  the  construction  and  the  equality  of  all  radii  of  the  same 
circle,  the  three  sides  GK^  KO,  and  OG  are  equal  respectively  to  a,  b,  c. 

Limitation.  It  is  necessary  that  any  two  of  the  sects 
should  be  greater  than  the  third,  or,  what  amounts  to  the  same, 
the  difference  of  any  two  sides  less  than  the  third.  For,  if 
a  and  c  were  together  less  than  b,  the  circles  in  the  figure  would 
not  meet ;  and,  if  they  were  together  equal  to  b,  the  point  K 


INEQUALITIES. 


47 


would  be  on  OG,  and  the  triangle  would  become  a  sect.  But 
as  we  have  proved,  in  156,  that  any  two  sides  of  a  triangle  are 
together  greater  than  the  third  side,  our  solution  of  Problem  VII 
will  apply  to  any  three  sects  that  could  possibly  form  a  triangle. 

163.  Corollary.  If  a  tri- 
angle be  made  of  hinged  rods, 
though  the  hinges  be  entirely 
free,  the  rods  cannot  turn  upon 
them  ;  the  triangle  is  rigid. 


Problem  VIII. 

164.  At  a  given  point  in  a  given  liney  to  make  an  angle  equal 
to  a  given  angle. 


Given,  the  point  A,  and  the  line  AB  and  4.  C, 
Required,  at  A,  with  arm  AB,  to  niake  an  angle  =  ^C 
Construction.    By  100,  join  any  two  points  in  the  arms  of  ^  C, 
thus  forming  a  A  DCE,     By  162,  make  a  triangle  whose  three  sides 
shall  be  equal  to  the  three  sides  oi  DCE]  making  the  sides  equal  to 
CD  and  CE  intersect  at  A. 

.-.     -2^  FAG  ^  i.  DCE, 
(zag.  Triangles  with  three  sides  respectively  equal  are  congruent.) 


48  THE  ELEMENTS  OF  GEOMETRY. 


CHAPTER  VI. 

PARALLELS. 

Theorem   XIX. 

^^S*  -^  t'^o  lines  cut  by  a  transversal  make  alternate  angles 
egualy  the  lines  are  parallel. 


/ 


7 


This  is  the  contranominal  of  142,  part  of  which  may  be 
stated  thus :  If  two  lines  which  meet  are  cut  by  a  transversal, 
their  alternate  angles  are  unequal. 

166.  Corollary.  If  two  lines  cut  by  a  transversal  make 
corresponding  angles  equal,  or  interior  or  exterior  angles  on 
the  same  side  of  the  transversal  supplemental,  the  lines  are 
parallel. 

For  we  know,  by  113,  that  either  of  these  suppositions 
makes  also  the  alternate  angles  equal. 

Exercises.  21.  From  a  given  point  without  a  given  line, 
draw  a  line  making  an  angle  with  the  given  line  equal  to  a 
given  angle. 


PARALLELS.  49 


Problem   IX. 

167.   Through  agivenpomtf  to  draw  a  line  parallel  to  a  given 
line. 


/-? 


Given,  a  line  AB  and  a  point  P. 

Required,  to  draw  a  line  through  P  ||  AB 

(using  II  for  the  words  "parallel  to"). 

Construction.     In  AB  take  any  point,  as  C.     By  100,  join  PC, 
At  P  in  the  Une  CP,  by  164,  make  4.  CPD  =  4.  PCB. 

PD  is  II  AB. 

Proof.  By  construction,  the  transversal  PC  makes  alternate  angles 
equal ;  4.  CPD  =  4  PCB, 

.',     by  165,  PD  is  II  AB. 

Exercises.  22.  Draw,  through  a  given  point  between  two^ 
lines  which  are  not  parallel,  a  sect,  which  shall  be  terminated 
by  the  given  lines,  and  bisected  at  the  given  point. 

23.  If  a  line  bisecting  the  exterior  angle  of  a  triangle  be- 
parallel  to  the  base,  show  that  the  triangle  is  isosceles. 

24.  Sects  from  the  middle  point  of  the  hypothenuse  of  a/, 
right-angled  triangle  to  the  three  vertices  are  equal. 

25.  Through  two  given  points  draw  two  lines,  forming,, 
with  a  given  line,  a  triangle  equiangular  with  a  given  triangle. 

26.  AC,  BD,  are  equal  sects,  drawn  from  the  extremities  of. 
the  sect  AB  on  opposite  sides  of  it,  such  that  the  angles  BAC„ 
ABD,  are  together  equal  to  a  straight  angle.  Prove  that  AB 
bisects  CD. 


50  THE  ELEMENTS  OF  GEOMETRY, 


Theorem   XX. 

1 68.  If  a  transversal  cuts  two  parallels y  the  alternate  angles 
are  equal. 


/L 


Hypothesis.    AB  H  CD  cuf  at  H  and  K  by  transversal  FG. 

Conclusion.     4.  HKD  =  ^  KHB. 

Proof.  A  line  through  H,  making  alternate  angles  equal,  is  paral- 
lel to  CD,  by  165. 

By  our  assumption,  99,  two  different  Imes  through  H  cannot  both 
be  parallel  to  CD ; 

.*•     by  31,  the  line  through  H  |I  CD  is  identical  with  the  line  which 
makes  alternate  angles  equal. 

But,  by  hypothesis,  AB  is  parallel  to  CD  through  H, 

.-.     i.  KHB  =  ^  HKD. 

169.  Corollary  I.  If  a  transversal  cuts  two  parallels,  it 
makes  the  alternate  angles  equal;  and  therefore,  by  113,  the 
corresponding  angles  are  equal,  and  the  two  interior  or  two 
exterior  angles  on  the  same  side  of  the  transversal  are  supple- 
mental. 

170.  Corollary  II.  If  a  line  be  perpendicular  to  one  of 
two  parallels,  it  will  be  perpendicular  to  the  other  also. 

171.  CoNTRANOMiNAL  OF  i68.  If  alternate  angles  are  un- 
equal, the  two  lines  meet. 


PARALLELS.  51 


So,  if  the  interior  angles  on  the  same  side  of  the  transversal 
are  not  supplemental,  the  two  lines  meet ;  and  as,  by  143,  they 
cannot  meet  on  that  side  of  the  transversal  where  the  two 
interior  angles  are  greater  than  a  straight  angle,  therefore  they 
must  meet  on  the  side  where  the  two  interior  angles  are  together 
less  than  a  straight  angle. 

172.  CoNTRANOMiNAL  OF  99.  Lines  in  the  same  plane  par- 
allel to  the  same  line  cannot  intersect,  and  so  are  parallel  to 
one  another. 


Theorem  XXI. 

173.  Each  exterior  angle  of  a  triafigle  is  equal  to  the  sum  of 
the  two  interior  opposite  angles. 


Hypothesis.     ABC  any  a,  wiih  side  AB  produced  to  D, 

Conclusion.     ^  CBD  =  ^A  -^  ^C. 

Proof.     From  B,  by  167,  draw  BF  \\  AC.    Then,  by  168,  ^  C 
^  CBF,  and  by  169,  :^  A  =  ^  DBF\ 

:.     by  adding,  4.  A  ^  t^C  =  4.  FBD  +  ^  CBF  =  4.  CBD. 


Exercises.  2y.  Each  angle  of  an  equilateral  triangle  is 
two-thirds  of  a  right  angle.  Hence  show  how  to  trisect  a  right 
angle. 

28.  If  any  of  the  angles  of  an  isosceles  triangle  be  two- 
thirds  of  a  right  angle,  the  triangle  must  be  equilateral. 


52  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XXII. 

174.    The  sum  of  the  three  interior  angles  of  a  triangle  is 
equal  to  a  straight  angle. 


Hypothesis.    ABC  any  A. 

Conclusion.     ^  CAB  +  4.  B  -^r  ^C=st.  ^. 

Proof.     Produce  a  side,  as  CAy  to  D. 

By  173,  tB  ^-tC  =  4  DAB. 

Add  to  both  sides  4.  BAC. 

.-.     4  CAB  +  ^^  +  ^C  =  ^  DAB  +  4.  BAC  =  st.  ^. 

175.   Corollary.      In  any  right-angled  triangle  the  two 
acute  angles  are  complemental. 


TRIANGLES.  53 


CHAPTER  VII. 
triangles. 

Theorem  XXIII. 

176.  Two  triangles  are  congruent  if  two  angles  and  an  oppO' 
site  side  in  the  one  are  equal  respectively  to  two  angles  and  the 
corresp07iding  side  in  the  other. 


Hypothesis.    ABC  and  DFG  As,  with 

^A=  tD, 

^C  =  ^G, 

AB  =  DF. 
Conclusion.     t^ABC^  t.  DFG, 

Proof.     Byi74,  2^^-f-^^H-^C'  =  st.  :^  =  ^  D  '\-  -4.  F  -\- 
^G, 

By  hypothesis,  ^  A  -[-  :^C  =  4.  D  -\-  ^Gy 

(89.  If  equals  be  taken  from  equals,  the  remainders  are  equal.) 

.-.     A  ABC  ^  A  DFG, 
(128.  Triangles  are  congruent  if  two  angles  and  the  included  side  are  equal  in  each.) 


54  THE  ELEMENTS  OF  GEOMETRY, 


Theorem  XXIV. 

"^11'  If  two  triangles  have  two  sides  of  the  one  equal  respec- 
tively to  two  sides  of  the  other^  and  the  angles  opposite  to  one 
pair  of  equal  sides  equals  theti  the  angles  opposite  to  the  other 
pair  of  equal  sides  are  either  equal  or  supplemental. 


The  angles  included  by  the  equal  sides  must  be  either  equal 
or  unequal. 

Case  I.     If  they  are  equal,  the  third  angles  are  equal. 

(174.  The  sum  of  the  angles  of  a  triangle  is  a  straight  angle.) 

Case  II.     If  the  angles  included  by  the  equal  sides  are 
unequal,  one  must  be  the  greater. 

Hypothesis.    ABC  and  FGH  as,  with 

¥^        =^^, 

AB     =      FG, 

BC     =      GH, 
^ABC  >  i.G. 

Conclusion.    ^  C  -f  ^  -^  =  st.  4^ . 
Proof.     By  164,  make  the  4.  ABD  =  :^Gy 

.-.     A  ABD  ^  A  FGH, 
(128.  Triangles  are  congruent  if  two  angles  and  the  included  side  are  equal  in  each.) 

.-.     ^  BDA  ^  tH> 
and 

BD     =      GH, 


TRIANGLES.  55 


But,  by  hypothesis,  BC  =^  GH, 

BC      =      BD, 

:,     i.  BDC  =  4.C. 
(126.  In  an  isosceles  triangle  the  angles  opposite  equal  sides  are  equal.) 

But  4.  BDA  +  t  BDC  =  St.  ^, 

/.     ^H  ^  4.C  ^  St.  ^. 

178.  Corollary  I.  If  two  triangles  have  two  sides  of  the 
one  respectively  equal  to  two  sides  of  the  other,  and  the  angles 
opposite  to  one  pair  of  equal  sides  equal,  then,  if  one  of  the 
angles  opposite  the  other  pair  of  equal  sides  is  a  right  angle,  or 
if  they  are  oblique  but  not  supplemental,  or  if  the  side  opposite 
the  given  angle  is  not  less  than  the  other  given  side,  the  tri- 
angles are  congruent. 

179.  Corollary  II.  Two  right-angled  triangles  are  con- 
gruent if  the  hypothenuse  and  one  side  of  the  one  are  equal 
respectively  to  the  hypothenuse  and  one  side  of  the  other. 


On  the  Conditions  of  Congruence  of  Two  Triangles. 

180.  A  triangle  has  three  sides  and  three  angles. 

The  three  angles  are  not  all  independent,  since,  whenever 
two  of  them  are  given,  the  third  may  be  determined  by  taking 
their  sum  from  a  straight  angle. 

In  four  cases  we  have  proved,  that,  if  three  independent 
parts  of  a  triangle  are  given,  the  other  parts  are  determined ; 
in  other  words,  that  there  is  only  one  triangle  having  those 
parts  :  — 

(124)  Two  sides  and  the  angle  between  them. 

(128)  Two  angles  and  the  side  between  them. 

(129)  The  three  sides. 

(176)  Two  angles  and  the  side  opposite  one  of  them. 

In  the  only  other  case,  two  sides  and  the  angle  opposite  one 


56  THE  ELEMENTS  OF  GEOMETRY. 

of  them,  if  the  side  opposite  the  given  angle  is  shorter  than 
the  other  given  side,  two  different  triangles  may  be  formed, 
each  of  which  will  have  the  given  parts.  This  is  called  the 
ambiguous  case. 

Suppose  that  the  side  a  and  side  r,  and  4.  C,  are  given.     If 
a>  c,  then,  making  the  2^  Cy  and  cutting  off  CB  =  a,  taking  B 


as  center,  and  describing  an  arc  with  radius  equal  to  c,  it  may 
cut  CD  in  two  points  ;  and  the  two  unequal  triangles  ABC 
and  ABC  will  satisfy  the  required  conditions. 

In  these  the  angles  opposite  the  side  BC  are  supplemental, 
by  177. 

Loci. 

181.  The  aggregate  of  all  points  and  only  those  points 
which  satisfy  a  given  condition,  is  called  the  Loctis  of  points 
satisfying  that  condition. 

182.  Hence,  in  order  that  an  aggregate  of  points,  L,  may  be 
properly  termed  the  locus  of  a  point  satisfying  an  assigned  con- 
dition, d  it  is  necessary  and  sufficient  to  demonstrate  the  fol- 
lowing pair  of  inverses  :  — 

(i)  If  a  point  satisfies  C,  it  is  upon  L. 

(2)  If  a  point  is  upon  L,  it  satisfies  C 

We  know  from  18,  that,  instead  of  (i),  we  may  prove  its 
contranominal :  — 

(3)  If  a  point  is  not  upon  Z,  it  does  not  satisfy  C 


TRIANGLES. 


57 


Also,  that,  instead  of   (2),  we  may  prove   the  obverse  of 

(I):- 

(4)  If  a  point  does  not  satisfy  C  it  is  not  upon  L, 

Theorem   XXV. 

183.  The  locus  of  the  point  to  which  sects  from  two  give7t 
points  are  equal  is  the  perpendicular  bisector  of  the  sect  joining 
them. 


T 

1 

\ 

\ 

, 

,' 

*\ 

i 

; 

C 

:b 

Hypothesis.    AB  a  sect,  C  its  mid-point 
PA  =  PB, 

Conclusion.    PC  J_  AB. 

Proof.    Join  PC. 

Then  A  ACP  is  ^  A  BCP, 

(129.  Triangles  with  the  three  sides  respectively  equal  are  congruent.) 

.-.     i.ACP  =  4.BCP, 

CP  is  the  perpendicular  bisector  of  AB. 

184.  Inverse. 

Hypothesis.    P,  any  point  on  CP  ±  AB  at  its  mid-point  C. 

Conclusion.    PA  =  PB. 

Proof,     t^  ACP  ^  aBCP. 

(124.  Two  triangles  are  congruent  if  two  sides  and  the  included  angle  are  equal  in 

each.) 


58 


THE  ELEMENTS  OF  GEOMETRY, 


Theorem   XXVI. 

185.  The  locus  of  points  from  which  perpendiculars  on  two 
given  intersecting  lines  are  equal,  consists  of  the  two  bisectors  of 
the  angles  betwee7i  the  given  lines. 


Hypothesis.     Given  AB  and  CD  intersecting  at  O,  and  P  any 
point  such  that  PM  1.  AB  ^  PN 1.  CD, 
Conclusion.     ^  POM  =  4.  PON. 
Proof,     a  POM  ^  A  PON, 
(179.  Right  triangles  having  hypothenuse  and  one  side  equal  are  congruent.) 

Therefore  all  points  from  which  perpendiculars  on  two  intersecting 
lines  are  equal,  lie  on  the  bisectors  of  the  angles  between  them. 

186.  Inverse. 

Hypothesis.    P,  any  point  on  bisector. 

Conclusion.    PM  =  PN. 
Proof,     a  POM  ^  PON. 

(176.  Triangles  having  two  angles  and  a  corresponding  side  equal  in  each  are 

congruent.) 

Hence,  by  110,  the  two  bisectors  of  the  four  angles  between  AB 
and  CD  are  the  locus  of  a  point  from  which  perpendiculars  on  AB  and 
CD  are  equal. 

Intersection   of  Loci. 

187.  Where  it  is  required  to  find  points  satisfying  two  con- 
ditions, if  we  leave  out  one  condition,  we  may  find  a  locus  of 
points  satisfying  the  other  condition. 


TRIANGLES.  59 


Thus,  for  each  condition  we  may  construct  the  correspond- 
ing locus ;  and,  if  these  two  loci  have  points  in  common,  these 
points,  and  these  only,  satisfy  both  conditions. 

Theorem  XXVII. 

188.  There  is  one^  and  only  one,  point  from  which  sects  to 
three  given  points  not  on  a  line  are  equal. 


o  ...-V ^ 


Hypothesis.     Given  A,  B,  and  C,  not  in  a  fine. 

Construction  and  Proof.  By  183,  every  point  to  which  sects  from 
A  and  B  are  equal,  lies  on  Z>iy,  the  perpendicular  bisector  of  AB. 
Again :  the  locus  of  points  to  which  sects  from  B  and  C  are  equal  is 
the  perpendicular  bisector  J^J^\ 

Now,  Z>Zy  and  FF'  intersect,  since,  if  they  were  parallel,  AB,  which 
is  perpendicular  to  one  of  them,  would  be  perpendicular  to  the  other  also, 
by  170,  and  ABC  would  be  one  line.  Let  them  intersect  in  O.  By 
95,  they  cannot  intersect  again. 

189.  Thus,  OB  =  OA, 

and 

OC  =  OA, 

.-.     OB  =  OC, 

Therefore,  by  183,  O  is  on  the  perpendicular  bisector  of  y^  C    There- 
fore 

Corollary.  The  three  perpendicular  bisectors  of  the  sides 
of  a  triangle  meet  in  a  point  from  which  sects  to  the  three  ver- 
tices of  the  triangle  are  equal. 


60  THE  ELEMENTS  OF  GEOMETRY. 


Problem   X. 

190.  To  Ji7td  points  from  which  perpendiculars  on  three  given 
lines  which  form  a  triangle  are  equal. 


Given,  a,  b,  r,  ihres  lines  inierseciing  in  the  three  disiinci 
points  Af  By  C. 

Solution.  The  locus  of  points  from  which  perpendiculars  on  the 
two  lines  a  and  b  are  equal  consists  of  the  two  bisectors  of  the  angles 
between  the  lines. 

The  locus  of  points  from  which  perpendiculars  on  the  two  lines  b 
and  c  are  equal,  similarly  consists  of  the  two  lines  bisecting  the  angles 
between  b  and  c.  Our  two  loci  consist  thus  of  two  pairs  of  Hnes.  Each 
line  of  one  pair  cuts  each  Hne  of  the  second  pair  in  one  point ;  so  that 
we  get  four  points,  O,  O^,  O2,  O^,  common  to  the  two  loci. 

191.  By  185,  the  four  intersection  points  of  the  bisectors  of  the 
angles  between  a  and  by  and  between  b  and  Cy  lie  on  the  bisectors  of 
the  angles  between  a  and  c.    Therefore 

Corollary.  The  six  bisectors  of  the  interior  and  exterior 
angles  of  a  triangle  meet  four  times,  by  threes,  in  a  point. 


POLYGONS.  6 1 


CHAPTER  VIII. 

POLYGONS. 

I.  Definitions. 

192.  A  Polygon  is  a  figure  formed  by  a  number  of  lines 
of  which  each  is  cut  by  the  following  one,  and  the  last  by  the 
first. 


193.  The  common  points    of   the    consecutive    lines    are 
called  the  Vertices  of  the  polygon. 

194.  The  sects  between  the  consecutive  vertices  are  called 
the  Sides  of  the  polygon. 


195.  The  sum  of  the  sides,  a  broken  line,  makes  the  Perim- 
eter of  the  polygon. 


62 


THE  ELEMENTS   OF  GEOMETRY. 


196.  The  angles  between  the  consecutive  sides  and  towards 
the  enclosed  surface  are  called  the  Interior  Angles  of  the  poly- 
gon.    Every  polygon  has  as  many  interior  angles  as  sides. 

197.  A  polygon  is  said  to  be  Convex  when  no  one  of  its 
interior  angles  is  reflex. 

198.  The  sects  joining  the  vertices  not  consecutive  are 
called  Diagonals  of  the  polygon. 


199.  When   the  sides   of  a  polygon  are  all  equal  to  one 
another,  it  is  called  Equilateral, 


200.  When  the  angles  of  a  polygon  are  all  equal   to   one 
another,  it  is  called  Equiangular, 


POL  YGONS.  63 


201.  A  polygon  which  is  both  equilateral  and  equiangular 
is  called  Regular, 


202.  Two  polygons  are  Mutually  Equilateral  if  the  sides  of 
the  one  are  equal  respectively  to  the  sides  of  the  other  taken 
m  the  same  order. 


203.  Two  polygons  are  Mutually  Equiangular  if  the  angles 
of  the  one  are  equal  respectively  to  the  angles  of  the  other 
taken  in  the  same  order. 


204.  Two   polygons  may  be  mutually  equiangular  without 
being  mutually  equilateral. 

205.  Except  in  the  case  of  triangles,  two  polygons  may  be 
mutually  equilateral  without  being  mutually  equiangular. 


64  THE  ELEMENTS  OF  GEOMETRY. 

'  206.  A  polygon  of  three  sides  is  a  Trigon  or  Triangle ;  one 
of  four  sides  is  a  Tetragon  or  Quadrilateral ;  one  of  five  sides 
is  a  Pentagon ;  one  of  six  sides  is  a  Hexagon;  one  of  seven 
sides  is  a  Heptagon ;  one  of  eight,  an  Octagon ;  of  nine,  a 
Nonagon ;  of  ten,  a  Decagon;  of  twelve,  a  Dodecagon ;  of  fif- 
teen, a  Quindecagon. 

207.  The  Surface  of  a  polygon  is  that  part   of  the   plane 
enclosed  by  its  perimeter. 

208.  A  Parallelogram    is   a  quadrilateral   whose   opposite 
sides  are  parallel. 


/ 


209.  A  Trapezoid  is  a  quadrilateral  with  two  sides  parallel. 


II.  General  Properties. 

Theorem  XXVIII. 

210.  If  two  polygons  be  mutually  equilateral  and  mutually 
equiangular^  they  are  congruent. 

Proof.  Superposition  :  they  may  be  applied,  the  one  to  the  other, 
so  as  to  coincide. 

Exercises.  29.  Is  a  parallelogram  a  trapezoid  ?  How 
could  a  triangle  be  considered  a  trapezoid  t 


POL  YGONS. 


65 


Theorem   XXIX. 

211.   The  sum  of  the  interior  angles  of  a  polygon  is  two  less 
straight  angles  than  it  has  sides. 


Hypothesis.    A  polygon  of  n  sides. 

Conclusion.    Sum  of  ^^s  =  (n  —  2)  st.  2^'s. 

Proof.     If  we  can  draw  all  the  diagonals  from  any  one  vertex  with 
out  cutting  the  perimeter,  then  we  have  a  triangle  for  every  side  of  the 
polygon,  except  the  two  which  make  our  chosen  vertex.    Thus,  we  have 
(«  —  2)  triangles,  whose  angles  make  the  interior  angles  of  the  polygon. 

But,  by  174,  the  sum  of  the  angles  in  each  triangle  is  a  straight 
angle, 

.'.     Sum  of  ^*s  in  polygon  =  (n  —  2)  st.  ^'s. 

212.  Corollary  I.     From  each  vertex  of  a  polygon  of  n 
sides  are  (u  —  3)  diagonals. 

213.  Corollary  II.     The  sum  of  the  angles  in  a  quadrilat- 
eral is  a  perigon. 


214 

gon  of  71  sides  is 


Corollary  III.     Each  angle  of  an  equiangular  poly- 

(n  —  2)  St.  ^'s 


66  THE  ELEMENTS  OF  GEOMETRY, 


Theorem   XXX. 

215.  In  a  convex  poly go7t^  the  sum  of  the  exterior  angles,  one 
at  each  vertex,  made  by  producing  each  side  in  order,  is  a  perigon. 


Hypothesis.    A   convex  polygon  of  n  sides,  each  in  order  pro- 
duced at  one  end. 

Conclusion.     Sum  of  exterior  2(^  's  =  perigon. 
Proof.     Every  interior  angle,  as  ^  A,  and  its   adjacent   exterior 
angle,  as  :^  x,  together  =  st.  ^ . 

.*.     all  interior  2^  's  +  all  exterior  ^  's  =  n  St.  ^  's. 
But,  by  211,  all  interior  ^'s  =  {n  —  2)  st.  ^'s. 

.*.     sum  of  all  exterior  2^'s  =  a  perigon. 

Exercises.     30.    How  many  diagonals  can  be  drawn  in  a 
polygon  of  n  sides  } 

31.  The  exterior  angle  of  a  regular  polygon  is  one-third  of 
a  right  angle :  find  the  number  of  sides  in  the  polygon. 

32.  The  four  bisectors  of  the  angles  of  a  quadrilateral  form 
a  second  quadrilateral  whose  opposite  angles  are  supplemental. 

33.  Divide  a  right-angled  triangle  into  two  isosceles  tri- 
angles. 

34.  In  a  right-angled  triangle,  the  sect  from  the  mid-point 
of  the  hypothenuse  to  the  right  angle  is  half  the  hypothenuse. 


POL  YGONS.  67 


III.  Parallelograms. 

Theorem   XXXI. 

216.  If  two  opposite  sides  of  a  quadrilateral  are  equal  and 
parallel^  it  is  a  parallelogram. 


Hypothesis.    ABCD  a  quadrilateral,  with  AB  =  and  \\   CD, 

Conclusion.    AD  \\  BC. 
Proof.    Join  AC.    Then  ^BAC  =  ^  A  CD. 
(168.  If  a  transversal  cuts  two  parallels,  the  alternate  angles  are  equal.) 

By  hypothesis,  AB  =  CD,  and  ^  C  is  common, 
/.     ^  BCA  =  ^  CAD, 
(124.  Triangles  are  congruent  if  two  sides  and  the  included  angle  are  equal  in  each.) 

.-.     BC  I!  AD. 
(165.  Lines  making  alternate  angles  equal  are  parallel.) 

Exercises.  35.  Find  the  number  of  elements  required  to 
determine  a  parallelogram. 

36.  The  four  sects  which  connect  the  mid-points  of  the 
consecutive  sides  of  any  quadrilateral,  form  a  parallelogram. 

37.  The  perpendiculars  let  fall  from  the  extremities  of  the 
base  of  an  isosceles  triangle  on  the  opposite  sides  will  include 
an  angle  supplemental  to  the  vertical  angle  of  the  triangle. 

38.  If  BE  bisects  the  angle  ^  of  a  triangle  ABC,  and  CE 
bisects  the  exterior  angle  A  CD,  the  angle  E  is  equal  to  one-half 
the  angle  A. 


6S  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XXXII. 

217.  The  opposite  sides  and  angles  of  a  parallelogram  are 
equal  to  one  another^  and  each  diagonal  bisects  it. 


Hypothesis.     Let  AC  be  a  diagonal  of  /z?  ABCD  (using  the  sign 
£y  for  the  word  "  parallelogram  "). 
Conclusions.    AB  =  CD. 
BC  =  DA. 
t  DAB  =  ^  BCD. 

^B  =:  4.D. 
A  ABC  ^  A  CDA. 
Proof.       ^  BAC  =  ^  ACD,  and 
^  CAD=  ■4.BCA) 
(168.  If  a  transversal  cuts  parallels,  the  alternate  angles  are  equal.) 

therefore,  adding, 

4.  BAC  +  4.  CAD  =  4.  BAD  =  4  ACD  +  4  ACB  =  4  BCD. 

Again,  the  side  A  C  included  between  the  equal  angles  is  common, 

.-.     aABC^  a  CDA, 
(128.  Triangles  are  congruent  if  two  angles  and  the  included  side  are  equal  in  each.) 

.-.    AB  =  CD,         BC  =  DA,         and  4  B  =  4  D. 

Exercises.  39.  If  one  angle  of  a  parallelogram  be  right,  all 
the  angles  are  right. 

40.  If  two  parallelograms  have  one  angle  of  the  one  equal 
to  one  angle  of  the  other,  the  parallelograms  are  mutually  equi- 
angular. 


POLYGONS. 


69 


218.  Corollary  I.     Any  pair  of  parallels  intercept  equal 
sects  of  parallel  transversals. 


219.  Corollary  II.  If  two  lines  be  respectively  parallel 
to  two  other  lines,  any  angle  made  by  the  first  pair  is  equal  or 
supplemental  to  any  angle  made  by  the  second  pair. 

220.  Corollary  III.  If  two  angles  have  their  arms  respec- 
tively perpendicular,  they  are  equal  or  supplemental. 


For,  revolving  one  of  the  angles  through  a  right  angle 
around  its  vertex,  its  arms  become  perpendicular  to  their  traces, 
and  therefore  parallel  to  the  arms  of  the  other  given  angle. 


70 


THE  ELEMENTS  OF  GEOMETRY. 


221.  Corollary  IV.  If  one  of  the  angles  of  a  parallelo- 
gram is  a  right  angle,  all  its  angles  are  right  angles,  and  it  is 
called  a  Rectangle, 


Corollary  V.    If  two  consecutive  sides  of  a  parallelo- 
equal,  all  its  sides  are  equal,   and  it  is  called  a 
Rhombus. 


gram  are 


223.  A  Square  is  a  rectangle  having  consecutive  sides  equal. 


224.  Both  diagonals,  AC SiVid  BD,  being  drawn,  it  may  with 
a  few  exceptions  be  proved  that  a  quadrilateral  which  has  any 
two  of  the  following  properties  will  also  have  the  others  : 

1.  AB  II  CD. 

2.  BC  II  DA.  C ^ ^777^ 

3.  AB  =  CD. 
BC  =  DA. 
i.  DAB  =  4.  BCD. 
4.  ABC  =  4  CDA. 
The  bisection  of  AC  hy  BD. 
The  bisection  of  BD  by  AC. 
The  bisection  of  the  njhy  AC. 
The  bisection  of  the  £17  by  BD. 


4. 

5- 
6. 

7- 

8. 

9- 
10. 


POLYGONS.  71 


These  ten  combined  in  pairs  will  give  forty-five  pairs ;  with 
each  of  these  pairs  it  may  be  required  to  establish  any  of  the 
eight  other  properties,  and  thus  three  hundred  and  sixty  ques- 
tions respecting  such  quadrilaterals  may  be  raised. 

For  example,  from  i  and  2,  217  proves  3, 

/.     A  ABE  ^  A  CDE, 

(128.  Two  triangles  are  congruent  if  two  angles  and  the  included  side  are  equal  in 

each.) 

.*.     the  diagonals  of  a  parallelogram  bisect  each  other. 

The  inverse  is  to  prove  i  and  2  from  7  and  8 : 
If  the  diagonals  of  a  quadrilateral  bisect  each  other,   the 
figure  is  a  parallelogram. 

225.  Since  by  170  two  lines  perpendicular  to  one  of  two 
parallels  are  perpendicular  to  the  other,  and  by  166  are  parallel, 
and  so  by  218  the  sects  intercepted  on  them  are  equal,  there- 
fore all  perpendicular  sects  between  two  parallels  are  equal. 


Exercises.     41.  If  the  mid-points  of  the  three  sides  of  a 
triangle  be  joined,  the  four  resulting  triangles  are  equal. 

42.  If  the  diagonals  of  a  parallelogram  be  equal,  the  paral- 
lelogram is  a  rectangle. 

43.  If  the  diagonals  of  a  parallelogram  cut  at  right  angles, 
it  is  a  rhombus. 

44.  If  the  diagonals  of  a  parallelogram  bisect  the  angles,  it 
is  a  rhombus. 

45.  If  the  diagonals  of  a  rectangle  cut  at  right  angles,  it  is 
a  square. 


72  THE  ELEMENTS  OF  GEOMETRY. 


Theorem    XXXIII. 

226.   The  three  perpendiculars  froin  the  vertices  of  a  triangle 
to  the  opposite  sides  ineet  in  a  point. 


Hypothesis.  Lei  Djy,  EE\  FF\  be  the  three  perpendiculars 
from  the  vertices  D,  E,  F,  io  the  opposite  sides  of  A  DEF. 

Conclusion.     Dlf ,  EE' ,  FF' ,  meet  in  a  point  O. 

Proof.  By  167,  through  D,  E,  F,  draw  AB,  BC,  CA  \\  FE,  DF, 
DE. 

Then  the  figures  ADFE,  DBFE,  are  parallelograms. 

By  217, 

.♦.    AD  =  FE  ^  DBy 

:.     Z>  is  mid-point  of  AB. 

In  same  way,  E  and  F  are  mid-points  oi  AC  and  BC. 

But,  since  DIf ,  EE,  FF',  are  respectively  ±  EF,  FD,  DEy 

.'.     they  must  he  ±AB,  AC,  and  BC, 
(170.  A  line  perpendicular  to  one  of  two  parallels  is  perpendicular  to  the  other.) 

.'.     by  189,  they  meet  in  a  point. 

Exercises.  46.  The  intersection  of  the  sects  joining  the 
mid-points  of  opposite  sides  of  any  quadrilateral  is  the  mid- 
point of  the  sect  joining  the  mid-points  of  the  diagonals. 


POL  YGONS. 


73 


Theorem   XXXIV. 

227.    If  three  or  more  parallels  intercept  equal  sects  on  one 
transversal^  they  intercept  equal  sects  on  every  transversal. 


Hypothesis.  AB  =  BC  =  CD  =,  etc.,  are  sects  on  the  trans- 
versa/ AC,  intercepted  by  the  parallels  a,  b,  c,  d,  etc. 

FGj  GHy  BKy  etc.,  are  corresponding  sects  on  any  other  trans- 
versal. 

Conclusion.    FG  =  GH  =  HK  =,  etc. 

Proof.  From  F,  G,  and  B,  draw,  by  167,  FL,  GM,  and  UN,  all 
parallel  to  AD ; 

.-.     they  are  all  equal,  because  AB  z=z  BC  =  CD  =,  etc. 
(2x7.  Opposite  sides  of  a  parallelogram  are  equal.) 

But  ^  GFL  =  4.  HGM  =  4.  KHN,  and 

4.  FGL  =  4  GHM  =  4  HKN, 
(169.  If  a  transversal  cuts  two  parallels,  the  corresponding  angles  are  equal.) 

/.     A  FGL  ^  A  GHM  ^  A  HKN, 

(176.  Triangles  are  congruent  if  they  have  two  angles  and  a  corresponding  side 

equal  in  each.) 

/.    FG=  GH  =:  HK. 

228.  Corollary.  The  intercepted  part  of  each  parallel  will 
differ  from  the  neighboring  intercepts  by  equal  sects. 


74 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XXXV. 

229.  The  line  drawn  through  the  mid-point  of  one  of  the  non- 
parallel  sides  of  a  trapezoid  parallel  to  the  parallel  sides^  bisects 
the  remaining  side. 


Hypothesis.     In  the  trapezoid  ABCD,  BF  =  FC,  and  AB  \\  CD 
WFG. 

Conclusion.    AG  ^  GD. 

Proof.    By  227,  if  parallels  intercept  equal  sects  on  any  transversal, 
they  intercept  equal  sects  on  every  transversal. 


230.  Corollary  I.  The  line  joining  the  mid-points  of  the 
non-parallel  sides  of  a  trapezoid  is  parallel  to  the  parallel  sides, 
for  we  have  just  proved  it  identical  with  a  line  drawn  parallel 
to  them  through  one  mid-point. 

231.  Corollary  II.  If  through  the  mid-point  of  one  side 
of  a  triangle  a  line  be  drawn  parallel  to  a  second  side,  it  will 
bisect  the  third  side ;  and,  inversely,  the  sect  joining  the  mid- 
points of  any  two  sides  of  a  triangle  is  parallel  to  the  third  side 
and  equal  to  half  of  it. 

232.  A  sect  from  any  vertex  of  a  triangle  to  the  mid-point 
of  the  opposite  side  is  called  a  Medial  of  the  triangle. 


POL  YGONS.  75 


233.  Three  or  more  lines  which  intersect  in  the  same  point 
are  said  to  be  Concurrent. 

234.  Three  or  more  points  which  lie  on  the  same  line  are 
said  to  be  Collinear, 


Theorem   XXXVI. 

235.   The  three  medials  of  a  triangle  are  concurrent  in  a 
trisection  point  of  each. 


Proof.  Let  two  medials,  AD  and  BE,  meet  in  O,  Take  G  the 
mid-point  of  OA,  and  Hoi  OB,    Join  GH,  HD,  DE,  EG. 

By  231,  in  A  ABO,  GH  is  equal  and  parallel  to  \AB  \  so  is  DE 
in  A  ABC', 

.*.    by  216,  GHDE  is  a  parallelogram. 
But,  by  224,  the  diagonals  of  a  parallelogram  bisect  each  other, 
/.    AG  ^  GO  ^  on,     and    BH  ^  HO  =  OE. 

So  any  medial  cuts  any  other  at  its  point  of  trisection  remote  from  its 
vertex, 

/,    the  three  are  concurrent. 


76 


THE  ELEMENTS  OF  GEOMETRY. 


236.  The  intersection  point  of  medials  is  called  the  Centroid 
of  the  triangle.  The  intersection  point  of  perpendiculars  from 
the  vertices  to  the  opposite  sides  is  called  the  Orthocenter  of  the 
triangle.  The  intersection  point  of  perpendiculars  erected  at 
the  mid-points  of  the  sides  is  called  the  Circumcenter  of  the 
triangle. 

IV.  Equivalence. 

237.  Two  plane  figures  are  called  equivalent  if  we  can  prove 
that  they  must  contain  the  same  extent  of  surface^  even  if  we  do 
not  show  how  to  cut  them  into  parts  congruent  in  pairs. 

238.  The  base  of  a  figure  is  that  one  of  its  sides  on  which 
we  imagine  it  to  rest. 

Any  side  of  a  figure  may  be  taken  as  its  base. 

239.  The  altitude  of  a  figure  is  the  perpendicular  from  its 
highest  point  to  the  line  of  its  base. 

So  the  altitude  of  a  parallelogram  is  the  perpendicular 
dropped  from  any  point  of  one  side  to  the  line  of  the  opposite 
side. 


240.  The  words  "  altitude  "  and  "  base  "  are  correlative ;  thus, 
a  triangle  may  have  three  different  altitudes. 


POL  YGONS. 


77 


Problem   XI. 
241.   To  describe  a  square  upon  a  given  sect. 


Given,  Me  SGci  AB. 

Required,  to  describe  a  square  on  AB. 

Construction.    By  135,  from  A  draw  AC  1.  AB. 

By  133,  in  AC  make  AB>  =  AB. 

By  167,  through  D  draw  DI^  ||  AB. 

By  167,  through  B  draw  BF  \\  AD. 
AF  will  be  the  required  square. 
Proof.     By  construction,  AF  is  a  parallelogram, 

AB  =      FD,     and        AD  =      BF, 
4.F     =  ^Ay       and    ^B     =  ^  D. 
(217.  In  a  parallelogram  the  opposite  sides  and  angles  are  equal.) 

But,  by  construction,  AB  =  AD, 

.'.    AF  is  equilateral. 

Again,  ^  A  -^  :i^  B  —  ?X,  ■^. 

(169.  If  two  parallel  lines  are  cut  by  a  transversal,  the  two  interior  angles  are 

supplemental.) 

But,  by  construction,  ^  A\s  rt., 

.-.     4.A  =  4.B, 
and  all  four  angles  are  rt. 


242.  Corollary. 
it  is  a  rectangle. 


If  a  parallelogram  have  one  angle  right, 


78 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XXXVII. 

243.  In  any  right-angled  triangle y  the  square  on  the  hypothe- 
nuse  is  equivalent  to  the  sum  of  the  squares  of  the  other  two 
sides. 


\3t 


ya 


Hypothesis,     a  ABC,  right-angled  at  B. 

Conclusion.     Square  on  AB  -\-  square  oTi  BC  =  square  on  AC. 

Proof.  By  241,  on  hypothenuse  A  C,  on  the  side  toward  the  A  ABC, 
describe  the  square  ADFC. 

On  the  greater  of  the  other  two  sides,  as  BC,  by  133,  lay  off  CG 
=  AB.    Join  FG. 

Then,  by  construction,  CA  =  FC,  and  AB  =  CG,  and  ^  CAB  = 
•4.  FCG,  since  each  is  the  complement  of  A  CB  \ 

.'.     A  ABC  ^  A  CGF. 

Translate  the  A  ABC  upward,  keeping  point  A  on  sect  AD,  and 
point  C  on  sect  CF,  until  A  is  on  D,  and  C  on  F;  then  call  B'  the 
position  of  B. 

Likewise  translate  CGF  to  the  right  until  C  is  on  A,  and  F  on  Z>  j 
then  call  G^  the  position  of  G. 


POL  YGONS. 


79 


The  resulting  figure,  AG'DB'FGBA^  will  be  the  squares  on  the 
other  two  sides,  AB  and  BC. 

For,  since  the  sum  of  the  angles  at  2)  =  st.  2>^, 

.*.     G'D  and  DB'  are  in  one  line. 
Produce  GB  to  meet  this  line  at  H. 

Then  GF  =  FB'  =  BC,     and    t\.  4.  B'  =  ^  GFB'  =  ^  FGH, 

.'.     GFB'H  equals  square  on  BC, 
Again,         G^'^  =  AB,     and    rt.  ^  G^'  =  ^  G'^^  =  ^  ^^^y, 

/.     AG'HB  is  the  square  on  ^^, 

.-.     sq.  oiAC=  sq.  of  ^^  +  sq.  of  ^C 

244.  Corollary.  Given  any  two  squares  placed  side  by 
side,  with  bases  AB  and  BC  in  line ;  to  cut  this  figure  into 
three  pieces,  two  of  which  being  translated  without  rotation, 
the  figure  shall  be  transformed  into  one  square. 


In  AC  take  AD  =  BC,  and  join  D  to  the  corners  of  the 
squares  opposite  B.  Two  right-angled  triangles  are  thus  pro- 
duced, with  hypothenuses  perpendicular  to  one  another.  Trans- 
late each  triangle  along  the  hypothenuse  of  the  other. 


80  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XXXVIII. 

245.   Two  triangles  are  congruent  if  they  are  mutually  equi- 
angular^  and  have  corresponding  altitudes  equal. 


Hypothesis.  As  ABC  and  BDF mutually  equiangular,  and  alti- 
tude BH  =  altitude  GB. 

Conclusion,    a  ABC  ^  A  BDF. 
Proof.    Rt.  a  BDG  ^  rt.  a  BCH, 
(176.  Triangles  are  congruent  when  two  angles  and  a  side  in  one  are  equal  to  two 
angles  and  a  corresponding  side  in  the  other.) 

BD    =      BC, 
.:     A  ABC  ^  A  BDF. 

(124.  Triangles  are  congruent  when  two  angles  and  the  included  side  are  equal  in 

each.) 

Exercises.  47.  If  from  the  vertex  of  any  triangle  a  per- 
pendicular be  drawn  to  the  base,  the  difference  of  the  squares 
on  the  two  sides  of  the  triangle  is  equal  to  the  difference  of  the 
squares  on  the  parts  of  the  base. 

48.  Show  how  to  find  a  square  triple  a  given  square. 

49.  Five  times  the  square  on  the  hypothenuse  of  a  right- 
angled  triangle  is  equivalent  to  four  times  the  sum  of  the 
squares  on  the  mcdials  to  the  other  two  sides. 


POL  YGONS. 


8l 


Theorem   XXXIX. 

246.  If  two  consecutive  sides  of  one  rectangle  he  respectively 
equal  to  two  consecutive  sides  of  anothery  the  rectangles  are  con- 
gruent. 

X     <?  € S 


Two  quadrilaterals  ABCD  and  FGHK,  wHh  all 


Hypothesis. 
^s  rt. 

AB  =  FG,  BC  =  GH, 

Conclusion.    ABCD  ^  FGHK. 
Proof.    AB  \\  CD,  and  AD  \\BC, 
(166.  K  interior  angles  on  same  side  of  transversal  are  supplemental,  lines  are 

parallel.) 

/.    a  rectangle  is  a  parallelogram, 

/.     AB  =  CD,     and    AD  =  BC, 
(217.  Opposite  sides  of  a  parallelogram  are  equal.) 

In  the  same  way,        FG  =  BK,    and     GH  =  FK, 

/.    ABCD  ^  FGHK, 
(2x0.  If  two  polygons  be  mutually  equilateral  and  mutually  equiangular,  they  are 

congruent.) 

247.  Corollary.     A  rectangle  is  completely  determined  by 
two  consecutive  sides ;  so  if  two  sects,  a  and  ^,  are  given,  we 


c& 


may  speak  of  the  rectangle  of  a  and  b,  or  we  may  call  it  the 
rectangle  ab.  It  can  be  constructed  by  the  method  given  in 
241  to  describe  a  square. 


82 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XL. 

248.  A  parallelogram  is  equivalent  to  the  rectangle  of  its  base 
and  altitude. 


C  jr       3 


Hypothesis.  ABCD  any  £=7,  of  which  side  AD  is  taken  as  base; 
AF  =  altitude  of  ^. 

Conclusion.    /17ABCD  =  rt.  /uA^FGD. 
Proof.    AF  ^  DG,  and  AB  =  DC. 

(217.  Opposite  sides  of  a  parallelogram  are  equal.) 

By  construction,  ^  G  and  2^  F  are  rt., 

.-.     aABF^^DCG. 

(179.  Right  triangles  are  congruent  if  hypothenuse  and  one  side  are  respectively 

equal  in  each.) 

From  the  trapezoid  ABGD  take  away  A  DCG,  and  then  is  left 
n?  ABCD.  From  the  same  trapezoid  take  the  equal  A  ABF^  and  there 
isleftthert.  zi^^i^G^/?. 

.-.    ^ABCD  =  rt.  ^  AFGD. 
(8g.  If  equals  be  taken  from  equals,  the  remainders  are  equal.) 

249.  Corollary.  All  parallelograms  having  equal  bases 
and  equal  altitudes  are  equivalent,  because  they  are  all  equiva- 
lent to  the  same  rectangle. 

Exercises.  50.  Equivalent  parallelograms  on  the  same 
base  and  on  the  same  side  of  it  are  between  the  same 
parallels. 


POL  YGONS.  83 


Exercises.  51.  Prove  248  for  the  case  when  (7  and  F  coin- 
cide. 

52.  If  through  the  vertices  of  a  triangle  lines  be  drawn 
parallel  to  the  opposite  sides,  and  produced  until  they  meet, 
the  resulting  figure  will  contain  three  equivalent  parallelo- 
grams. 

53.  On  the  same  base  and  between  the  same  parallels  as  a 
given  parallelogram,  construct  a  rhombus  equivalent  to  the 
parallelogram. 

54.  Divide  a  given  parallelogram  into  two  equivalent  paral- 
lelograms. 

55.  Of  two  parallelograms  between  the  same  parallels,  that 
is  the  greater  which  stands  on  the  greater  base.  Prove  also  an 
inverse  of  this. 

56.  Equivalent  parallelograms  situated  between  the  same 
parallels  have  equal  bases. 

57.  Of  parallelograms  on  equal  bases,  that  is  the  greater 
which  has  the  greater  altitude. 

58.  A  trapezoid  is  equivalent  to  a  rectangle  whose  base  is 
half  the  sum  of  the  two  parallel  sides,  and  whose  altitude  is  the 
perpendicular  between  them. 

59.  The  sect  joining  the  mid-points  of  the  non-parallel  sides 
of  a  trapezoid  is  half  their  sum. 

60.  If  E  and  F  are  the  mid-points  of  the  opposite  sides,  AD, 
BC,  of  a  parallelogram  ABCD,  the  lines  BE,  DF,  trisect  the 
diagonal  AC. 

61.  Any  line  drawn  through  the  intersection  of  the  diag- 
onals of  a  parallelogram  to  meet  the  sides,  bisects  the  surface. 

62.  The  squares  described  on  the  two  diagonals  of  a  rhom- 
bus are  together  equivalent  to  the  squares  on  the  four  sides. 

63.  Bisect  a  given  parallelogram  by  a  line  passing  through 
any  given  point. 

64.  In  244,  what  two  rotations  might  be  substituted  for  the 
two  translations  ? 


84 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XLI. 

Alternative   Proof. 

250.  All  parallelogrmns  having  equal  bases  and  equal  alti- 
tudes are  equivalent. 

2)> zrC 


Hypothesis.     T\no  ^s  with  equal  bases  and  equal  aliiiudes. 

Conclusion.     They  can  be  cut  into  parts  congruent  in  pairs. 

Construction.  Place  the  parallelograms  on  opposite  sides  of  their 
coincident  equal  bases,  AB.  Produce  a  side,  as  FB,  which  when  con- 
tinued will  enter  the  other  parallelogram.  If  it  cuts  out  of  the  parallel- 
ogram at  H  before  reaching  the  side  CD  opposite  AB,  then  will  the 
other  cutting  side,  as  CB,  when  produced,  also  leave  the  ^  ABFG 
before  reaching  the  side  FG  opposite  AB ;  that  is,  the  point,  K,  where 
CB  cuts  the  line  A  G  will  be  on  the  sect  A  G. 

For,  through  H  and  K  draw  BL  and  KM  \\  AB, 

Then,  in  As  ABB  and  ABK 

^1  =  ^1,     ^2  =  ^2, 
(i68.  If  a  transversal  cuts  two  parallels,  the  alternate  angles  are  equal.) 

and  side  AB  is  common ; 

.-.     A  ABB  ^  A  ABK, 


POL  YGONS.  85 


But  altitude  of  A  AHB  <  altitude  of  £jABCD, 

/.    altitude  of  A  ABK  <  altitude  of  £y  ABFG,  and  K  is  on  sect  A  G. 

Now,  A  ABH  ^  A  BHL,  and  A  ABK  ^  A  BKM, 

(217.  The  diagonal  of  a  parallelogram  makes  two  congruent  triangles.) 

Taking  away  these  four  congruent  triangles,  we  have  left  two  paral- 
lelograms, HLCD  and  KMFG,  with  equal  bases,  and^  equal  but  dimin- 
ished altitudes.  Treat  these  in  the  same  way  as  the  parallelograms  first 
given  ;  and  so  continue  until  a  produced  side,  as  FRQ,2irvd  so  the  other, 
CRN,  also,  reaches  the  side  opposite  the  base  before  leaving  the  paral- 
lelogram. 

Then,  as  before,  the  As  FRN  and  QRC  are  mutually  equiangular; 
but  now  we  know  their  corresponding  altitudes  are  equal. 

.'.     by  241,  they  are  congruent, 
.-.    FN  =.  CQ, 
.-.    GN  =  DQ, 

Also,         QR  ^  RF  =  PG,     and    NR  ^  RC  ^  PD, 
4.G        =  ^  FRP  =  4.  RQD, 
4.D       =^4.  CRP  =  4.  RNG, 
4.  DPR  =  4.  PRN, 
4GPR  =  4.PRQ', 

and  therefore  the  remaining  trapezoids,  PRQD  ^  PRNG. 

(210.   If  two  polygons  be  mutually  equilateral  and  mutually  equiangular,  they 

are  congruent.) 

251.  Corollary.  Since  a  rectangle  is  a  parallelogram, 
therefore  a  parallelogram  is  equivalent  to  the  rectangle  of  its 
base  and  altitude. 

Exercises.  65.  How  do  you  know  that  HLCD  and  KMFG 
have  equal  altitudes  "i 

66.  How  do  you  know,  that,  if  Q  is  on  the  sect  CD,  N  is 
on  the  sect  FG  ? 


S6  THE  ELEMENTS  OF  GEOMETRY. 


Theorem    XLII. 

252.  A  triangle  is  equivalent  to  half  the  rectangle  of  its  base 
and  altitude. 


Hypothesis,     a  ABC,  with  base  AC  and  altiiude  BD. 

Conclusion,     a  AB C  —  half  the  rectangle  oi  AC  and  BD. 
Proof.     Through  A  draw  AF  \\  CB,  and  through  B  draw  BF  \\  CA, 
meeting  AF  in  F; 

.'.     L^ACB  ^  ^BFA. 
(217.  The  diagonal  of  a  parallelogram  bisects  it.) 

But  £jAFBC  =  rectangle  oi  AC  and  BD, 

(248.  A  parallelogram  is  equivalent  to  the  rectangle  of  its  base  and  altitude.) 

.-.     t.ABC=^\  ^AFBC  =  J  rectangle  oi  AC  and  BD. 

253.  Corollary  I.  All  triangles  on  the  same  base  having 
their  vertices  in  the  same  line  parallel  to  the  base,  are  equivalent. 

254.  Corollary  II.  Triangles  having  their  vertices  in  the 
same  point,  and  for  their  bases  equal  sects  of  the  same  line, 
are  equivalent. 

255.  Corollary  III.  If  a  parallelogram  and  a  triangle  be 
upon  the  same  base  and  between  the  same  parallels,  the  paral- 
lelogram is  double  the  triangle. 

Exercises.  6j.  Equivalent  triangles  on  equal  bases  have 
equal  altitudes. 


POL  YGONS.  Zj 


Theorem   XLIII. 

256.  If  through  any  point  on  the  diagonal  of  a  parallelogram 
two  lines  be  drawn  parallel  to  the  sideSy  the  two  parallelograms y 
one  on  each  side  of  the  diagonal y  will  be  equivalejit. 


Hypothesis.  P  any  point  on  diagonal  BD  of  /=;  ABCD\  FG 
and  HK  lines  ih rough  P  ||  AB  and  BC  respectively,  and  meeting 
the  four  sides  in  the  points  F,  G,  Hy  K, 

Conclusion.     ^AKPG=^PFCH. 
Proof,    a  ABD  =  a  BCD. 

A  KBP  =  A  BFP, 

A  GPD  =  A  PBD. 
(217.  The  diagonal  of  a  parallelogram  bisects  it.) 

From  A  ABD  take  away  A  KBP  and  A  GPD,  and  we  have  lisft 
C7  AKPG.  From  the  equal  A  BCD  take  away  the  equal  As  BFP  and 
PHDy  and  we  have  left  the  £=7  PFCH, 

.-.    ^AKPG=^PFCH, 
(89.  If  equals  be  taken  from  equals,  the  remainders  will  be  equal.) 

257.  In  figures  like  the  preceding,  parallelograms  like  KBFP 
and  GPHD  are  called  parallelograms  about  the  diagonal  BD ; 
while  Z3^s  AKPG  and  PFCH  are  called  complements  of  par- 
allelograms about  the  diagonal  BD. 

Exercises.  68.  When  are  the  complements  of  the  paral- 
lelograms about  a  diagonal  of  a  parallelogram  congruent  ? 


88  THE  ELEMENTS  OF  GEOMETRY. 

V.  Problems. 

Problem   XII. 

258.   To  describe  a  parallelogram  equivalent  to  a  given  tri- 
angle, and  having  an  angle  equal  to  a  given  angle. 


M  JU 


Given,  a  ABC  and  ^  G. 

Required,  to  describe  a  parallelogram  —  A  ABC,  while  having  an 
angle  =  2^  G. 

Construction.     Bisect  ACmD. 
At  D,  by  164,  make  ^  ADF  =  4.  G, 
By  167,  through  A  draw  AH  \\  DF, 
By  167,  through  B  draw  BFH  \\  CA. 
AHFD  will  be  the  parallelogram  required. 
Proof.    Join  DB.    Then  A  ABD  =  aBCD, 

(254.  Triangles  having  their  vertices  in  same  point,  and  for  bases  equal  sects  of 
the  same  line,  are  equivalent.) 

/.     A  ABC  is  double  A  AB£>. 

But  /I7AHFD  is  also  double  A  ABD, 

(255.  If  a  parallelogram  and  a  triangle  be  upon  the  same  base  and  between  the 
same  parallels,  the  parallelogram  is  double  the  triangle.) 

/.    npAHFD  =  A  ABC, 
and,  by  construction,  ^  ADF  =  ^  G, 


POLYGONS. 


89 


Problem  XIII. 

259.  On  a  given  sect  as  base,  to  describe  a  parallelogram 
equivalent  to  a  given  triangle^  and  having  an  angle  equal  to  a 
given  angle, 

X ^^ Q  /V 


Given,  Mo  seci  AB,  a  CDF,  4.  G. 

Required,  to  describe  on  AB  a  parallelogram  =  A  CDF,  while  having 
an  angle  =  ^  G. 

Construction.    By  258,  make  ^  BHKL  —  A  CDF,  and  having 
^  HBL  =  7^  G,  and  place  it  so  that  BH  is  on  line  AB  produced. 
Produce  KL.     Draw  AM  \\  BL.    Join  MB. 

4.  HKM  +  i-  KMA  =  St.  ^, 
(169.  If  a  transversal  cuts  two  parallels,  it  makes  the  two  interior  angles  supple- 
mental.) 

/.     ^HKM  ^  4KMB<%\..4., 
Therefore  KH  and  MB  meet  if  produced  through  H  and  B. 
(171.  If  a  transversal  cuts  two  lines,  and  the  interior  angles  are  not  supplemental, 

the  lines  meet.) 

Let  them  meet  in  Q.     Through  Q  draw  QN  ||  HA,  and  produce 
LB  and  MA  to  meet  QN'm  P  and  N. 

:.    ^  NPBA  =  C7  BHLK, 
(256.  Complements  of  parallelograms  about  the  diagonal  are  equivalent.) 
/.    £yNPBA  =  A  CFD,        and         4.  ABP  =  4  G. 

260.  Corollary.     Thus  we  can  describe  on  a  given  base  a 
rectangle  equivalent  to  a  given  triangle. 


90  THE  ELEMENTS  OF  GEOMETRY. 

Problem    XIV. 

261.   To  describe  a  triangle  equivalent  to  a  given  polygon. 
J) 


Given,  a  polygon  ABCDFG, 

Required,  to  constntct  an  equivalent  triangle. 

Construction.  Join  the  ends  of  any  pair  of  adjacent  sides,  as  AB 
and  BC,  by  the  sect  CA. 

Through  the  intermediate  vertex,  B,  draw  a  hne  H  CA^  meeting  GA 
produced  in  H.     Join  HC. 

Polygon  ABCDFGA  =  polygon  BCDFGB, 
and  we  have  obtained  an  equivalent  polygon  having  fewer  sides. 

Proof.   /\ABC=  a  ABC. 

(253.  Triangles  having  the  same  base  and  equal  altitudes  are  equivalent.) 

Add  to  each  the  polygon  A  CDFGA, 

.'.     ABCDFGA  =  BCDFGB. 
In  the  same  way  the  number  of  sides  may  be  still  further  diminished 
by  one  until  reduced  to  three. 

262.  Corollary  I.  Hence  we  can  describe  on  a  given 
base  a  parallelogram  equivalent  to  a  given  polygon,  and  having 
an  angle  equal  to  a  given  angle. 

263.  Corollary  II.  Thus  we  can  describe  on  a  given 
base  a  rectangle  equivalent  to  a  given  polygon. 

264.  Remark.  To  compare  the  surfaces  of  different  poly- 
gons, we  need  only  to  construct  rectangles  equivalent  to  the 
given  polygons,  and  all  on  the  same  base. 

Then,  by  comparing  the  altitudes,  we  are  enabled  to  judge 
of  the  surfaces. 


POL  YGONS. 


91 


VI.  Axial   and   Central    Symmetry. 

265.  If  two  figures  coincide,  every  point  A  in  the  one  coin- 
cides with  a  point  A'  in  the  other.  These  points  are  said  to 
correspond. 

Hence  to  every  point  in  one  of  two  congruent  figures  there 
corresponds  one,  and  only  one,  point  in  the  other ;  those  points 
being  called  "  corresponding  "  which  coincide  if  one  of  the  two 
figures  is  superimposed  upon  the  other.  Hence,  calling  those 
parts  corresponding  which  coincide  if  the  whole  figures  are 
made  to  coincide,  it  follows,  that  corresponding  parts  of  con- 
gruent figures  are  themselves  congruent. 


Symmetry  with  Regard  to  an  Axis. 

266.  If  we  start  with  two  figures  in  the  position  of  coinci- 
dence, and  take  in  the  common  plane  any  line  or,  we  may  turn 
the  plane  of  one  figure  about  this  line  x  until  its  plane,  after 
half  a  revolution,  coincides  again  with  the  plane  of  the  other 
figure. 


The  two  figures  themselves  will  then  have  distinct  positions 
in  the  same  plane ;  but  they  will  have  this  property,  that  they 


THE  ELEMENTS   OF  GEOMETRY. 


can  be  made  to  coincide  by  folding  the  plane  over  along  the 
line  X, 

Two  figures  in  the  same  plane  which  have  this  property  are 
said  to  be  symmetrical  with  regard  to  the  line  x  as  an  axis  of 
symmetry. 


Symmetry  with  Regard  to  a  Center. 

267.  If  we  take  in  the  common  plane  of  two  coincident 
figures  any  point  Xy  we  may  turn  the  one  figure  about  this 
point  so  that  its  plane  slides  over  the  plane  of  the  other  figure 
without  ever  separating  from  it. 

Let  this  turning  be  continued  until  one  line  to  Jf,  and 
therefore  the  whole  moving  figure,  has  been  turned  through  a 
straight  angle  about  X. 

Then  the  two  congruent  figures  still  lie  in  the  same  plane, 
and  have  such  positions  that  one  can  be  made  to  coincide  with 
the  other  by  turning  it  in  the  plane  through  a  straight  angle 
about  the  fixed  point  X. 

Two  figures  which  have  this  property  are  said  to  be  sym- 
metrical with  regard  to  the  point  X  as  center  of  symmetry. 


268.  Any  single  figure  has  axial  symmetry  when  it  can  be 
divided  by  an  axis  into  two  figures  symmetrical  with  respect  to 
that  axis,  or  has  central  symmetry  when  it  has  a  center  such 
that  every  line  drawn  through  it  cuts  the  figure  in  two  points 
symmetrical  with  respect  to  this  center. 


POL  YGONS. 


93 


Theorem   XLIV. 

269.  If  a  figure  has  two  axes  of  symmetry  perpendiciUar  to 
each  othery  then  their  intersection  is  a  center  of  symmetry. 


: 

; 

^^. 

_\;>1 

.'""'"  : 

-V-l^ 

- 

>\ 

-       ^ 

'^i 

J 

For,  if  X  and  y  be  two  axes  at  right  angles,  then  to  a  point  A  will 
correspond  a  point  A'  with  regard  to  x  as  axis. 

To  these  will  correspond  points  A^  and  A/  with  regard  to  j  as  axis. 
These  points  A^  and  A/,  will  correspond  to  each  other  with  regard  to  x. 
To  see  this,  let  us  first  fold  over  along  y ;  then  A  falls  on  A,y  and  A' 
on  A/. 

If  we  now,  without  folding  back,  fold  over  along  x.  A,  and  with  it 
Ai,  will  fall  on  A\  which  coincides  \vith  A^. 

At  the  same  time  OA  and  OA^  coincide,  so  that  the  angles  A  Ox 
and  AiOx'  are  equal,  where  x'  denotes  the  continuation  of  ^  beyond  O. 
It  follows,  that  AOAl  are  in  a  line,  and  that  the  sect  AA^  is  bisected  at 
O,  or  (P  is  a  center  of  symmetry  for  AAlj  and  similarly  for  Aj_  and  A' , 


BOOK   11. 


RECTANGLES. 

270.  A  Contimious  Aggregate  is  an  assemblage  in  which  two 
adjacent  parts  have  the  same  boundary. 

271.  A  Discrete  or  Discontinuous  Aggregate  is  one  in  which 
two  adjacent  parts  have  different  boundaries. 

A  pile  of  cannon  balls  is  a  discrete  aggregate.  We  know 
that  any  adjacent  two  could  be  painted  different  colors,  and  so 
they  have  direct  independent  boundaries. 

Our  fingers  are  a  discontinuous  aggregate. 

272.  All  counting  belongs  first  to  the  fingers. 

273.  There  is  implied  and  bound  up  in  the  word  "  number  " 
the  assumption  that  a  group  of  things  comes  ultimately  to  the 
same  finger,  in  whatever  order  they  are  counted. 

This  proposition  is  involved  in  the  meaning  of  the  phrase 
*•  distinct  things." 

Any  one  and  any  other  of  them  make  two.  If  they  are 
attached  to  two  of  my  fingers  in  a  certain  order,  they  can  also 
be  attached  to  the  same  fingers  in  the  other  order.  Thus,  one 
order  of  a  group  of  three  distinct  things  can  be  changed  into 
any  other  order  while  using  the  same  fingers,  and  so  on  with  a 
group  of  four,  etc. 

95 


96  THE  ELEMENTS  OF  GEOMETRY. 

274.  By  generalizing  the  use  of  the  fingers  in  counting, 
man  has  made  for  himself  a  counting  apparatus,  which  each 
one  carries  around  in  his  mind.  This  counting  apparatus,  the 
natural  series  of  numbers,  was  made  from  a  discrete  aggregate, 
and  so  will  only  correspond  exactly  to  discrete  aggregates. 

275.  In  a  row  of  shot,  we  can  find  between  any  two,  only  a 
finite  number  of  others,  and  sometimes  none  at  all. 

Just  so  in  regard  to  any  two  numbers.  A  row  of  six  shot 
can  be  divided  into  two  equal  parts ;  but  the  half,  which  is 
three,  we  cannot  divide  into  two  equal  parts :  and  so  in  a  series 
of  numbers. 

276.  But  in  136  we  have  shown  how  any  sect  whatever  may 
be  bisected,  and  the  bisection  point  is  the  boundary  of  both 
parts.  So  a  line  is  not  a  discrete  aggregate  of  points.  It  is 
something  totally  different  in  kind  from  the  natural  series  of 
numbers. 

277.  The  science  of  numbers  is  founded  on  the  hypothesis 
of  the  distinctness  of  things.  The  science  of  space  is  founded 
on  the  entirely  different  hypothesis  of  continuity. 

278.  Numbers  are  essentially  discontinuous,  and  therefore 
unsuited  to  express  the  relations  of  continuous  magnitudes. 

279.  In  arithmetic  we  are  taught  to  add  and  multiply  num- 
bers :  we  will  now  show  how  the  laws  for  the  addition  and 
multiplication  of  these  discrete  aggregates  are  applicable  to 
sects,  which  are  continuous  aggregates. 


THE   COMMUTATIVE   LAW   FOR   ADDITION. 

280.  In  a  sum  of   numbers  we    may  change   the   order  in 
which  the  numbers  are  added. 

If  X  and  y  represent  numbers,  this  law  is  expressed  by  the 

equation 

X  -\-  y  =  y  ■\'  X, 


RECTANGLES.  g/ 


It  depends  entirely  on  the  interchangeability  of  any  pair 
of  the  units  of  numeration. 

281.  The  sum  of  two  sects  is  the  sect  obtained  by  placing 
them  on  the  same  line  so  as  not  to  overlap,  with  one  end  point 
in  common. 


Thus,  the  sum  of  the  sects  a  and  b  means  the  sect  AC^ 
which  can  be  divided  into  two  parts, 

AB  =  a,        and        BC  =  b, 

282.  The  commutative  law  holds  for  the  summation  of  sects. 
a  ■\-  b  =  b  -^  a. 


M 


AC=  CA') 


for  AC  revolved  through  a  straight  angle  may  be  superimposed 
upon  C^A\  and  will  coincide  point  for  point. 

The  more  general  case,  where  three  or  more  sects  are 
added,  follows  from  a  repetition  of  the  above. 

Thus,  the  commutative  law  for  addition  in  geometry  depends 
entirely  on  the  possibility  of  motion  without  deformation.    . 


98 


THE  ELEMENTS  OF  GEOMETRY. 


283.  The  sum  of  two  rectangles  is  the  hexagon  formed  by 
superimposing  two  sides,  and  bringing  the  bases  into  the  same 
line. 


Thus,  if  two  adjacent  sides  of  one  are  a  and  b,  and  of 
the  other  c  and  d^  the  sum  of  the  rectangles  ab  and  cd  is 
ABCDFGA. 

284.  The  commutative  law  holds  for  the  addition  of  rect- 
angles ;  that  is,  the  sum  is  independent  of  the  order  of  summa- 
tion. 


ab  •\-  cd  ^  cd  -\-  ab\ 

for    ABCDFGA    turned    over    may    be    superimposed    upon 
AB'CD'FCA\  and  will  coincide  with  it. 

Since,  by  263,  we  can  describe  on  a  given  base  a  rectangle 
equivalent  to  a  given  polygon,  the  more  general  case,  where 
three  or  more  rectangles  are  added,  follows  from  a  repetition 
of  the  above. 


RECTANGLES. 


99 


THE   ASSOCIATIVE   LAW. 

285.  In  getting  a  sum  of  numbers,  we  may  add  the  num- 
bers together  in  groups,  and  then  add  these  groups. 

If  we  use  parentheses  to  mean  that  the  terms  enclosed  have 
been  added  together  before  they  are  added  to  another  term, 
this  law  may  be  expressed  symbolically  by  the  equation 

X  -{■  {y  ■\-  z)  =  X  ^  y  -\-  z. 

286.  The  associative  law  holds  for  the  summation  of  sects. 

a  +  {3+c)  =  a  +  d-\-c  =  AD. 


J£  2 
<x 


ty 


C 

I  1 


C       2) 


o. 


287.  The  associative  law  holds  for  the  summation  of  rect- 
angles. 

ad  +  {bf  +  eg)  =  ad  +  bf  -^  eg. 


f 


9 


JS. 


lOO 


THE  ELEMENTS  OF  GEOMETRY. 


THE   COMMUTATIVE    LAW    FOR    MULTIPLICATION. 

288.  The   product   of    numbers   remains   unaltered   if    the 
factors  be  interchanged. 

xy  =  yx. 

289.  The  commutative  law  holds  for  the  rectangle  of  two     | 
sects.  I 


a. 


or 


If  a  and  b  are  any  two  sects,  rectangle  ab  =  rectangle  ba, 

ab  =  ba, 


for  rectangle  ab  may  be  so  applied  to  rectangle  ba  as  to  coin- 
cide with  it. 


THE   DISTRIBUTIVE   LAW. 

290.  To  multiply  a  sum  of  numbers  by  a  number,  we  may 

multiply  each   term   of  the   sum,  and   add  the  products  thus 

obtained. 

x{^y  ■\-  z)  —  xy  ■\-  xz. 

291.  The  distributive  law  holds  when  for  numbers  and  prod- 
ucts we  substitute  sects  and  rectangles. 


a{b  -{■  c)  —  ab  -\-  ac 'y 


^     (^tuuu^ 


RECTANGLES. 


lOI 


for  if  we  add  the  rectangle  ab  to  the  rectangle  aCy  so  that 
a  side  a  in  the  one  shall  coincide  with  an  equal  side  a  in  the 
other,  the  sum  makes  a  rectangle  whose  base  is  ^  +  ^  and 
whose  altitude  is  a ;  that  is,  the  rectangle  a{p  +  c). 

In  the  same  way,  by  adding  three  rectangli.s .  <?£;  the  sanje  , 

altitude,  we  get  ,•,""''*    '  '       "•>'' » 

a{b  -\-  c  ■\-  a)—  ab  -^  ac  ^  adfj\  f,,  \  >'"{''  *'*  ;'.; 


We  may  state  this  in  words  as  follows : 

If  there  be  any  two  sects  one  of  which  is  divided  into  any 
number  of  partSy  the  rectaitgle  contained  by  the  two  sects  is 
equivalent  to  the  recta^tgles  contained  by  the  undivided  sect  and 
the  several  parts  of  the  divided  sect. 

292.  \i  b  '\-  c  =.  a,  then 

ab  -^  ac  ■=.  a{b  +  c)  =  aa  ==  a^. 


Therefore 

If  a  sect  be  divided  into  any  two  parts^  the  rectangles  con- 
tained by  the  whole  and  each  of  the  parts  are  together  equivalent 
to  the  square  on  the  whole  sect. 


102 


THE  ELEMENTS  OF  GEOMETRY. 


293.  If  r  =  ^,  then 

a{b  -^  c)  —  a{b  -\-  a)  —  ab  ■\-  aa  ^=-  ab  -\-  a^. 


amC 


Therefore 

If  a  sect  he  divided  into  any  two  parts^  the  rectangle  con- 
tained by  the  whole  and  one  of  the  parts  is  equivalent  to  the 
rectangle  contained  by  the  two  parts ^  together  with  the  square  on 
the  aforesaid  part. 

294.  The  rectangle  of  two  equal  sects  is  a  square,  and 
{a  +  by  is  only  a  condensed  way  of  writing  {a  +  b){a  -\-  b). 

But,  by  the  distributive  law, 

{a  +  b){a  +  b)  =  {a  -\-  b)a  -\-  {a  +  b)b. 
By  the  commutative  law, 

(a  +  b)a  =  a{a  +  b). 
By  the  distributive  law, 

a{a  -\-  b)  =  aa  -\-  ab  =  a^  -\-  ab. 


In  the  same  way, 


^  g 


(a  +  b)b  =  ab  -\-  b^, 

{a  +  by  =  a^  +  ab  -^  {ab  +  b^). 


RECTANGLES. 


103 


By  the  associative  law, 

a^  -ir  ab  -^  {ab  +  <^)  =  ^'  +  {ab  +  ab)  +  b^  =  a^  -^  2ab  +  b', 

,',     {a  +  by  =  a"  -\-  2ab  +  ^. 
Therefore 

If  a  sect  be  divided  into  a7iy  two  parts,  the  square  on  the 
whole  sect  is  equivalent  to  the  squares  on  the  two  parts y  together 
with  twice  the  rectangle  contained  by  the  two  parts. 

295.  Corollary.  The  square  on  a  sect  is  four  times  the 
square  on  half  the  sect. 


a 


:-»■   «. 


c    z.     J)  a,  :b 


I 


296.  By  the  distributive  law,  and  294, 

a{a  -{-  b  -\-  b)  -\-  b^  =  a""  -h  ab  -^  ab  +  b^  =  (a  ■}-  b)'. 

From  this,  if  we  bisect  the  sect  AB  at  C,  and  divide  it  into 
two  unequal  parts  at  D,  and  for  BD  put  a,  and  for  DC  put  b, 
we  get  the  theorem  : 

If  a  sect  is  divided  into  two  equal  parts,  and  also  into  two 
unequal  parts,  the  rectangle  contained  by  the  uneqtial  parts, 
together  with  the  square  oji  the  line  between  the  points  of  section, 
is  equivalent  to  the  square  on  half  the  sect. 

297.  If  CD  =  b,  and  AC  =  a  +  b,  then  their  sum  AD  =  a 
•}-  b  -\-  b,  and  their  difference  is  a ;  therefore  the  rectangle  con- 
tained by  their  sum  and  difference  equals  a(a  +  2b),  which,  by 
296,  is  the  difference  between  (a  +  by  and  b^. 


104- 


THE  ELEMENTS  OF  GEOMETRY. 


Therefore 

The  rectangle  contained  by  the  sum  and  difference  of  any  two 
sects  is  equivalent  to  the  difference  between  the  squares  on  those 
sects. 

298.  If  we  bisect  the  sect  AB  at  C,  and  produce  it  to  Z?, 
and  for  BD  put  a,  and  for  BC  put  3,  then  the  above  equation, 
a{a  ■\-  b  '\-  b)  -\-  b"^  =i  {a  •\-  by,  gives  us  the  theorem : 

If  a  sect  be  bisected^  and.produced  to  any  pointy  the  rectangle 
contained  by  the  whole  line  thus  produced  and  the  part  of  it  pro- 
duced, together  with  the  square  on  half  the  sect  bisected,  is  equiv- 
alent to  the  square  on  the  line  which  is  made  up  of  the  half  and 
the  part  produced. 


299.  By  294, 

{a  +  by  -{-  a^  =  a^  +  2ab  +  ^^  +  cC, 

By  the  associative  law, 

a"  ^  2ab  -^  b"  -^  a^  =^  20"  +  2ab  +  ^^ 

By  the  distributive  law, 

2a^  4-  2ab  -{-  b^  =  2a{a  -^  b)  -\-  b*. 

By  the  commutative  law, 

2a(a  -\-  b)  -{-  b''  =  2{a  +  b")a  +  ^, 

/.     {a  -\-  by  -^  a^  =  2(a  -{-  b)a  +  b^. 

Therefore 

If  a  sect  be  divided  into  any  two  parts,  the  squares  on  the 
whole  line  and  on  one  of  the  parts  are  equivalent  to  twice  the 


RECTANGLES. 


•105 


rectangle  co7ttained  by  the  whole  and  that  party  together  with 
the  square  on  the  other  part. 


Z-  « 


300.  By  the  commutative  and  distributive  laws,  and  294, 

4(«  +  b)a  +  b^  =  4a^  +  4ab  +  ^^  =  (2a  +  by. 
By  the  associative  and  commutative  laws, 

(2a  -{-by=^(a  +  a  +  by  =  (la  +  ^]  +  «)% 
•     .-.    4{a  4-  b)a  4-  ^^  =  {[a  4-  ^]  +  ^y. 
a       4t ^ 


^ 


Therefore 

7/"  a  sect  be  divided  into  any  two  parts,  four  times  the  rect- 
angle contained  by  the  whole  sect  and  one  of  the  parts,  together 


a 
a 

I 


I06  THE  ELEMENTS  OF  GEOMETRY. 

with  the  square  on  the  other  part  ^  is  equivalent  to  the  square  on 
the  line  which  is  made  up  of  the  whole  sect  and  the  first  part. 
301.  By  the  associative  and  commutative  laws,  and  294, 

{\_a  -f  /5]  +  ay  +  ^^2  =  (<5  4-  ^ay  +  ^' 

=  /^  4-  /i^ab  +  4^^  +  ^*  =  20"  +  A,ab  +  2<^  +  2«^ 

By  the  distributive  law,  and  294, 

20"  4-  Adb  +  2b'-  4-  2a^  —  2{a^  -\-  2ab  +  b^)  ■\-  20"  =  2{a  -\-  by  +  2<a!% 

.-.     {[a  4-  ^]  4-  ay  -\-  b^  —  2{a  4-  by  +  20", 


,       ^""^^       »    "    .      ^       . 


Therefore 

If  a  sect  be  divided  into  two  equal  and  also  two  unequal 
partSy  the  squares  on  the  two  unequal  parts  are  together  double 
the  squares  on  half  the  sect  and  on  the  line  between  the  points 
of  section. 

302.  By  294,  and  the  associative  and  distributive  laws, 

(dz  +  ^)2  +  ^  =  «^  4-  2ab  +  b^  +  b"" 

=  J  +   ^IV   2ab  +    2bA   =    2ffl   +  2  (^    +   bj. 

Therefore 

If  a  sect  be  bisected  and  produced  to  any  pointy  the  square  on 
the  whole  line  thus  produced  and  the  square  on  the  part  of  it 
produced  are  together  double  the  squares  on  half  the  sect  and  on 
the  line  made  up  of  the  half  and  the  part  produced. 

303.  The  projection  of  a  point  on  a  line  is  the  foot  of  the 
perpendicular  from  the  point  to  the  line. 


(TU, 


V--JL- — -^M" 


•^-^^,.,    ir:x. 


a 


c, 


RECTANGLES. 


107 


304.  The  projection  of  a  sect  upon  a  line  is  the  part  between 
the  perpendiculars  dropped  upon  the  line  from  the  ends  of  the 
sect. 


For  example,  A'B'  is  the  projection  of  the  sect  AB  on  the 
line  c. 

Theorem  I. 

305.  In  an  obtuse-angled  triangle y  the  square  on  the  side  oppo- 
site the  obtuse  angle  is  greater  than  the  sum  of  the  squares  on 
the  other  two  sides  by  twice  the  rectangle  co7itained  by  either  of 
those  sides  and  the  projection  of  the  other  side  upon  it. 


Hypothesis,     a  ABC,  wiih  ^  CAB  obtuse. 

Conclusion.    ^^  =  ^  +  ^  +  2bf 
Proof.    By  294, 

(b  +  J'Y  =  3^  +  2hj  +  j\ 
Adding  h^  to  both  sides, 

{b  +  JY  J^  h-  =  b^-\-  2bj  +  y-  +  h\ 
But 

{b  +  j'Y  -\-  h^  =  a",         and       j^  -\-  h^  z=  c", 

(243.  In  a  right  triangle,  the  square  of  the  hypothenuse  equals  the  sum  of  the 

squares  of  the  other  sides.) 

.-.      ^2  _  ^2  _|_  2^-   ^  ^. 


i- 


io8 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   II. 

306.  In  any  triangle^  the  square  on  a  side  opposite  any  acute 
angle  is  less  than  the  sum  of  the  squares  on  the  other  two  sides 
by  twice  the  rectangle  contained  by  either  of  those  sides  and  the 
projection  of  the  other  side  upon  it. 


Hypothesis,     a  ABC,  with  ^  C  acute. 

Conclusion,    c^  -{-  2bJ  =  a^  -{-  b^. 

Proof.    By  295,  

b''  +  j^  =  2b/  +  Alf, 
Adding  k^  to  both  sides, 

b"^  +  J^  +  y^"  =  2b/  +  A&  +  /i\ 

But  

y^  4.  ^2  ^  «2^        and        Alf  +  h^  =  c^, 

(243.  In  a  right  triangle,  the  square  of  the  hypothenuse  equals  the  sum  of  the 
squares  of  the  other  sides.) 

.*.     /^^  +  «2  =  2bj  +  c^, 

307.  Having  now  proved  that  in  a  triangle, 

By243,  if  ;^^  =  rt.  2^,  .'.  a' =  b- -{- c- ; 

By  305,  if  ^A  >  rt.  ^,  .'.  a^  >  b^  •{-  c^ ; 

By  306,  if  ^A  <  rt.  ^,  /.  a^  <  b^  +  c^. 
Therefore,  by  33,  Rule  of  Inversion, 

If  a""  =  b^  -{-  f%  .-.  ^  ^  ==  rt.  ^  j 

If  a'  >  b^  +  ^,  .-.  ^A>  rt.  ^  ; 

If  a^  <  b^  +  ^,  /.  ^A<  rt.  i. 


RECTANGLES.  109 


Theorem   III, 

308.  In  any  triangle,  if  a  medial  be  drawn  from  the  vertex 
to  base^  the  sum  of  the  sqicares  on  the  two  sides  is  equivalent  to 
twice  the  square  on  half  the  base,  together  with  twice  the  square 
on  the  medial. 


Hypothesis,     a  ABC,  with  medial  BM  =  i. 

Conclusion,    a"^  +  c^  =.  2{^by  +  21^. 

Proof.  — Case  I.     If  ^  BMA  =  ^  BMC,  then 

a^  =  {\by  +  /%         and         c^  =  {^by  +  /^ 

(243.  In  a  right  triangle,  the  square  of  the  hypothenuse  equals  the  sum  of  the 
squares  of  the  other  sides.) 

Case  II.  If  ^  BMA  does  not  equal  ^  BMC,  one  of  them  must 
be  the  greater.     Call  the  greater  BMC. 

Then,  in  the  obtuse-angled  triangle  BMC,  by  305, 

a-^  {^dy  +  i^+  2{\bj). 
In  A  BMA,  by  306, 

^  +  2(|^y)  =  {iby  +  i\ 
Adding, 

«-  +  ^  4-  2{\bj)  =  2{\by  +  2/«  +  2{\bj), 

,\     a^  +  c"  =  2{iby  4-  2i^. 

309.  Corollary.  The  difference  of  the  squares  on  the 
two  sides  is  equivalent  to  twice  the  rectangle  of  the  base  and 
the  projection  of  the  medial  on  it. 


no 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   IV. 

310.  The  sum  of  the  squares  on  the  four  sides  of  any  quad- 
rilateral is  greater  than  the  sum  of  the  squares  on  the  diagonals 
by  four  times  the  square  on  the  sect  joining  the  mid-points  of  the 
diagonals.  ^  -q 


Hypothesis.    EF  is  the  sect  joining  fhe  mid-points  of  the  diag- 
onals AC  and  BD  of  the  quadrilateral  ABCD. 

Conclusion.    ~A&  +  'BC^  +  C^'  -f  'DA' 

=  ^C"  +  BL^  -\-4£F\ 
Proof.    Draw  BE  and  DE. 

By  308,  _         ,.r\2         

AB"  +  BC'  =  2\—-]  +  2^^, 

and 

C&-h^'=  2{^  +  2DE'. 
Adding, 

AB" -^  BC^ -\-  CD" -\- DA"  =^  ^(^T  +  ^^^  +  ^^^' 
But,  by  308, 

BE'+  DE'^:  4^y  +  2KF^, 

=  AC""  +  B&  +  4^^^- 


RECTANGLES. 


Ill 


311.  Corollary.  The  sum  of  the  squares  on  the  four 
sides  of  a  parallelogram  is  equivalent  to  the  sum  of  the  squares 
on  the  diagonals,  because  the  diagonals  of  a  parallelogram 
bisect  each  other. 


Problem   I. 
312.   To  square  atty  polygon. 


.  H 


r^- 


Given,  any  polygon  N. 

Required,  to  describe  a  square  equivalent  to  N. 

Construction.     Describe,  by  263,  the  rectangle  ABCD  =  N. 

Then,  if  AB  —  BC^  the  required  square  is  ABCD. 

If  AB  be  not  equal  to  BC,  produce  BA^  and  cut  off  AF  =  AD, 
Bisect  BF  in  G,  and,  with  center  G  and  radius  GB,  describe  FLB. 

Produce  DA  to  meet  the  circle  in  H. 

The  square  on  AH  shall  be  equal  to  N. 

Proof.    Join  GH.    Then,  by  296,  because  the  sect  BF  is  divided 
equally  in  G  and  unequally  in  A^ 


rectangle  BA,AF  +  AC^  ^  GF"  ^  GH""  =  AH""  +  AC^, 


by  243 ; 


AH^  =  rectangle  BA,AF  =  rectangle  BA,AD  =  N. 


112 


THE  ELEMENTS  OF  GEOMETRY. 


Problem   II. 

313.  To  divide  a  given  sect  into  two  parts  so  that  the  rect- 
angle contained  by  the  whole  and  one  of  the  parts  shall  be  equal 
to  the  square  on  the  other  part, 

Jfr ,X 


Given,  ihe  sect  AB. 

Required,  to  find  a  point  C  such  that  rectangle  AB,BC  =  AC*. 

Construction.     On  AB^  by  241,  describe  the  square  ABDF, 

By  136,  bisect  AF  in  G.    Join  BG, 

Produce  FA,  and  make  GH  =  GB. 

On  AH  describe  the  square  AHKC. 

Then  AB,BC  =  ZC^ 

Proof.  Produce  KC  to  Z.  Then,  by  298,  because  FA  is  bisected 
in  G  and  produced  to  H,  rectangle  FH,BA  +  'A&  =  'GH^  =  'GB" 
=  GA'  +  AB\  

.-.    FB,HA  =  AB", 
But,  since  HK  =  HA, 

.-.    FH,HK  =  AB", 
Take  from  each  the  common  part  AL, 

.',    HC==  CD; 

that  is, 

AC^  =  CB,BD  =  CB,BA, 


RECTANGLES.  1 13 


Problem   III. 


314.    To   describe  an   isosceles  triangle  having  each  of  the 
angles  at  the  base  double  the  ajigle  at  the  vertex. 


Construction.  Take  any  sect,  AB,  and,  by  313,  divide  it  in  Cso 
that  rectangle  AB,AC  =  ~^^.  With  center  C  and  radius  CB,  describe 
a  circle. 

With  same  radius,  but  center  A,  describe  a  circle  intersecting  the 
preceding  circle  in  D.    Join  AD,  BD,  CD. 

ABD  will  be  the  triangle  required. 

Proof.     By  construction,  CB  =  CD  =  DA, 

By  134,  bisect  ^  ADC  by  DF.  By  137,  DF  \%  Jl  AC,  and  bisects 
AC. 

Then  'AB'  -f  Z^  =  'BD'  +  2AB,AF. 

(306.  The  square  on  a  side  opposite  an  acute  angle  is  less  than  the  squares  on  the 
other  two  sides  by  twice  the  rectangle  of  either  and  the  projection  of  the  other 
on  it.) 

But  ^C  =  2AFy  .'.    AB,AC  ==  AB,2AF  -  2AB,AF\ 

.'.     by  our  initial  construction,  BC^  ==  2AB,AF, 
.-.     AB"  +  AD"  =  Bif  +  BC\ 
But,  by  construction,  AD  ^  BC,     .'.    AB"  =  BD',     .-.   AB  =  BD, 
.'.     ^  ADB  =  ^  DAB  =  ^  ACD  =  ^  B  -^  ^  BDC  =  2  ^  B. 

(173.  The  exterior  angle  of  a  triangle  equals  the  sum  of  the  two  opposite  interior 

angles.) 


BOOK    III. 


THE   CIRCLE. 

I.   Primary   Properties. 

315.  If  a  sect  turns  about  one  of  its  end  points,  the  other 
end  point  describes  a  curve  called  the  Circle. 

316.  The  fixed  end  point  is  called  the  Center  of  the  circle. 

317.  The  moving  sect  in  any  position  is  called  a  Radius  of 
the  circle. 

318.  As  the  motion  of  a  sect  does  not  enlarge  or  diminish 
it,  all  radii  are  equal. 

319.  Since  the  moving  sect,  after  revolving  through  a  peri- 
gon,  returns  to  its  original  position,  therefore  the  moving  end 
point  describes  a  closed  curve. 

This  divides  the  plane  into  two  surfaces,  one  of  which  is 
swept  over  by  the  moving  sect.  This  finite  plane  surface  is 
called  the  surface  of  the  circle. 

Any  part  of  the  circle  is  called  an  Arc, 


SS5 


Il6  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   I. 

320.  The  sect  to  a  point  fivm  the  ce^iter  of  a  circle  is  less 
tha?t,  equal  to,  or  greater  than,  the  radius,  according  as  the  point 
is  within,  on,  or  without  the  circle. 


Proof.     If  a  point  is  on  the  circle,  the  sect  drawn  to  it  from  the 
center  is  a  radius,  for  it  is  one  of  the  positions  of  the  describing  sect. 
Any  point,  Q,  within  the  circle  lies  on  some  radius,  OQR, 


/.     OQ  <  OR. 

If  ^S*  is  without  the  circle,  then  the  sect  OS  contains  a  radius  OR, 
:.     OS  >  OR. 

321.  By  33,  Rule  of  Inversion,  a  point  is  within,  on,  or  with- 
out the  circle  according  as  its  sect  from  the  center  is  less  than, 
equal  to,  or  greater  than,  the  radius. 

322.  A  Secant  is  a  line  which  passes  through  two  points  on 
the  circle. 


THE   CIRCLE.  11/ 


Theqrem    II. 
323.  A  secant  can  meet  the  circle  in  only  two  points. 


Proof.  By  definition,  all  sects  joining  the  center  to  points  on  the 
circle  are  equal,  but  from  a  point  to  a  line  there  can  be  only  two  equal 
sects. 

(155.  No  more  than  two  equal  sects  can  be  drawn  from  a  point  to  a  line.) 

324.  A  Chord  is  the  part  of  a  secant  between  the  two  points 
where  it  intersects  the  circle. 

325.  A  Segment  of  a  circle  is  the  figure  made  by  a  chord 
and  one  of  the  two  arcs  into  which  the  chord  divides  the  circle. 

326.  When  two  arcs  together  make 
an  entire  circle,  each  is  said  to  be  the 
Expleinejit  of  the  other. 

327.  When  two  explemental  arcs  are 
equal,  each  is  a  Semicircle. 

328.  When  two  explemental  arcs  are 
unequal,  the  lesser  is  called  the  Minor 
Arc,  and  the  greater  is  called  the  Major 
Arc. 

329.  A  segment  is  called  a  Major  or  Minor  Segment  ac- 
cording as  its  arc  is  a  major  or  minor  arc. 


Il8  THE  ELEMENTS  OF  GEOMETRY. 

Theorem   III. 
330.  A  circle  has  only  one  center. 


Hypothesis.    Lei  F,  G,  and  H  be  points  on  a  O. 

Conclusion.     O  FGH  has  only  one  center. 

Proof.     Join  FG  and  GH. 

Since,  by  definition,  a  center  is  a  point  from  which  all  sects  to  the 
circle  are  equal,  therefore  any  center  of  a  circle  through  F  and  G  is  in 
the  perpendicular  bisector  of  FG,  and  any  center  of  a  circle  through 
G  and  H  is  in  the  perpendicular  bisector  of  GH. 

(183.  The  locus  of  the  point  to  which  sects  from  two  given  points  are  equal  is  the 
perpendicular  bisector  of  the  sect  joining  them.) 

But  these  two  perpendicular  bisectors  can  intersect  in  only  one 
point, 

/.     O  FGH  has  only  one  center. 

331.  Corollary  I.  The  perpendicular  bisector  of  any 
chord  passes  through  the  center. 

332.  Corollary  II.  To  find  the  center  of  any  given  cir- 
cle, or  of  any  given  arc  of  a  circle,  draw  two  non-parallel 
chords  and  their  perpendicular  bisectors.  The  center  is  the 
point  where  these  bisectors  intersect. 

333.  A  Diameter  is  a  chord  through  the  center. 

334.  A  diameter  is  equal  to  two  radii :  so  all  diameters  are 
bisected  by  the  center  of  the  circle,  and  are  equal. 


THE   CIRCLE.  II9 


Theorem   IV. 
335.  Circles  of  equal  radii  are  congruent. 


Hypothesis.  Two  circles  of  which  C  and  O  are  the  centers, 
and  radius  CD  =  radius  OP. 

Conclusion.     The  circles  are  congruent. 

Proof.  Apply  one  circle  to  the  other  so  that  the  center  O  shall 
coincide  with  center  C,  and  sect  OF  fall  upon  line  CD.  Then,  because 
OF  =  CD,  the  point  F  will  coincide  with  the  point  D.  Then  every 
particular  point  in  the  one  circle  must  coincide  with  some  point  in  the 
other  circle,  because  of  the  equality  of  radii. 

(321.  A  point  is  on  the  circle  when  its  sect  from  the  center  is  equal  to  the  radius.) 

/.      0  C  ^  O  6>. 

336.  Corollary.  After  being  applied,  as  above,  the  second 
circle  may  be  turned  about  its  center ;  and  still  it  will  coincide 
with  the  first,  though  the  point  P  no  longer  falls  upon  D. 

Hence,  considering  one  circle  as  the  trace  of  the  other,  — 

A  circle  can  be  made  to  slide  along  itself  by  being  turned 
about  its  center. 

This  fundamental  property  of  this  curve  allows  us  to  turn 
any  figure  connected  with  the  circle  about  the  center  without 
changing  its  relation  to  the  circle. 


120  THE  ELEMENTS  OF  GEOMETRY. 

337.  Circles  which   have  the  same  center  are  called  Con- 
centric. 


Theorem  V. 

338.  Different  concentric  circles  cannot  have  a  point  in  com- 
mon. 


Proof.  The  points  of  the  circle  with  the  lesser  radius  are  all  within 
the  larger  circle. 

(321.  A  point  is  within  the  circle  if  its  distance  from  the  center  is  less  than  the 

radius.) 

339.  First  Contranominal  of  338.  Two  different  circles 
with  a  point  in  common  are  not  concentric. 

340.  Second  Contranominal  of  338.  Two  concentric  cir- 
cles with  a  point  in  common  coincide. 

341.  The  center  of  a  circle  is  a  Center  of  Symmetry^  the 
end  points  of  any  diameter  being  corresponding  points. 

This  follows  at  once  from  the  definition  of  Central  Sym- 
metry, and  the  fundamental  property  that  the  circle  slides 
along  itself  when  turned  about  its  center,  and  so  coincides 
with  itself  after  turning  about  the  center  through  any  angle. 
The  circle  is  the  only  closed  curve  which  will  slide  upon  its 
trace. 


THE   CIRCLE.  121 


Theorem  VI. 

342.  The  perpendicular  from  the  center  of  a  circle  to  a  secant 
bisects  the  chord ;  and,  if  a  line  thivngh  the  center  bisect  a  chord 
not  passing  through  the  center^  it  cuts,  it  at  right  angles. 


Proof.  For  any  chord,  there  is  only  one  perpendicular  front  the 
center,  only  one  line  through  the  mid-point  and  center,  only  one  per- 
pendicular bisector;  and  these,  by  331,  are  identical. 

(331.  The  perpendicular  bisector  of  any  chord  passes  through  the  center.) 

343.  Every  diafneter  is  an  Axis  of  Symmetry, 

For  if  we  fold  over  along  a  diameter,  every  point  on  the 
part  of  the  circle  turned  over  must  fall  on  some  point  on 
the  other  part,  since  its  sect  from  the  center,  which  remains 
fixed,  is  a  radius. 

Inverse.  Every  line  which  is  an  axis  of  symmetry  of  a 
circle  contains  a  diameter. 

For,  if  not,  there  would  be  a  point  symmetrical  to  the 
center,  and  this,  too,  would  again  be  a  center  :  the  circle  would 
thus  have  two  centers. 

(330.  A  circle  has  only  one  center.) 

Each  circle  has  therefore,  besides  its  center  of  symmetry^ 
an  infinite  number  of  axes  of  symmetry. 


122  THE  ELEMENTS  OF  GEOMETRY. 

Theorem  VII. 
344.  Every  chord  lies  wholly  within  the  circle. 


Hypothesis.     Lei  A  and  B  be  any  two  points  in  0  ABC. 

Conclusion.  Every  point  on  chord  AB  between  A  and  B  is  within 
the  O  ABC. 

Proof.    Take  any  point,  B>,  in  chord  AB. 

By  332,  find  O,  the  center  of  the  circle. 
.      Join  OA,  0Z>,  OB. 

Then  OB  makes  a  greater  sect  than  OD  from  the  foot  of  the  per- 
pendicular from  O  on  the  line  AB, 

(342.  The  perpendicular  from  center  on  line  AB  bisects  chord  AB.) 

.-.     OB  >  OD, 

(r54.  The  oblique  which  makes  the  greater  sect  from  the  foot  of  the  perpendicular 

is  the  greater.) 

/.    D  is  within  the  circle. 
.<320.  A  point  is  within  the  circle  if  its  sect  from  the  center  is  less  than  the  radius.) 

345'  Corollary.  If  a  line  has  a  point  within  a  circle,  it  is 
a  secant,  for  the  radius  is  greater  than  the  perpendicular  from 
the  center  to  this  line :  so  there  will  be  two  sects  from  the 
center  to  the  line,  each  equal  to  the  radius ;  that  is,  the  line 
will  pass  through  two  points  on  the  circle. 

Thus,  again,  the  circle  is  a  closed  curve. 


THE   CIRCLE.  1 23 


Theorem  VIII. 

346.  In  a  circle,  two  chords  which  are  not  both  diameters  do 
not  mutually  bisect  each  other. 


Hypothesis.  Lei  ihe  chords  AB,  CD,  which  do  not  both  pass 
through  the  center,  cut  one  another  in  the  point  F,  in  the  O  A  CBD. 

Conclusion.     AB  and  CD  do  not  mutually  bisect  each  other. 

Proof.  If  one  of  them  pass  through  the  center,  it  is  not  bisected 
by  the  other,  which  does  not  pass  through  the  center. 

If  neither  pass  through  the  center,  find  the  center  (9,  and  join  OF, 

If  F  is  the  bisection  point  of  one  of  the  chords,  as  AB,  then 

4.  OFB  =  rt.  4., 

(342.  If  a  line  through  the  center  bisect  a  chord  not  passing  through  the  center,  it 
cuts  it  at  right  angles.) 

.*.     "4-  OFD  is  oblique, 

.-.     OF  does  not  bisect  CD. 

Exercises.  69.  What  is  the  locus  of  mid  points  of  parallel 
chords } 

70.  Prove  by  symmetry  that  the  diameter  perpendicular  to 
a  chord  bisects  that  chord,  bisects  the  two  arcs  into  which  this 
chord  divides  the  circle,  and  bisects  the  angles  at  the  center 
subtended  by  these  arcs. 


124 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   IX. 

347.  If  from  any  point  not  the  center^  sects  he  drawn  to  the 
different  points  of  a  circle,  the  greatest  is  that  which  meets 
the  circle  after  passing  through  the  center ;  the  least  is  part  of  the 
same  line. 


Hypothesis.    From  any  point  A,  sects  are  drawn  to  a  Q  DBHK, 
whose  center  is  C. 

First  Conclusion.    A  CB  >  AD. 

Proof.    Join  CD. 

Then,  because  CB  =  CD, 

.:    AB  =  AC  +  CB  =  AC  -j-  CD  >  AD. 

(156.  Any  two  sides  of  a  triangle  are  greater  than  the  third.) 

Second  Conclusion.    AH  <  AK. 
Proof.     When  A  is  within  the  circle, 

BC  =  KC  <KA  +  AC,  by  156. 
Taking  away  A  C  from  both  sides,  .*.     AH  <  AK. 

When  A  is  on  the  circle,  AH  is  a  point. 
When  A  is  without  the  circle, 

AC  =  AH  -{-  HC  <AK  ■\-  KC,  by  156. 
Taking  away  HC  =  KC,  .\    AH  <  AK. 

348.  Corollary.     The  diameter  of  a  circle  is  greater  than 
any  other  chord. 


THE   CIRCLE.  1 25 


Theorem   X. 

349*  If  from  any  point  three  sects  drawn  to  a  circle  are  equal, 
that  point  is  the  center. 


Hypothesis.     From  ihe  point  O  to  Q  ABC,  lei  OA  =  OB  =  OC. 

Conclusion.     O  is  center  of  O  ABC. 
Proof.    Join  AB,  and  BC. 

O  is  on  the  perpendicular  bisector  of  chord  AB,  and  also  on  that 
of  chord  BC, 

(183.  The  locus  of  a  point  to  which  sects  from  two  given  points  are  equal  is  the 
perpendicular  bisector  of  the  sect  joining  them.) 

.-.     O  is  center  of  O  ABC. 

(332.  The  center  is  the  intersection  of  perpendicular  bisectors  of  two  non-parallel 

chords.) 

350.  CoNTRANOMiNAL  OF  349.  From  any  point  not  the 
center,  there  cannot  be  drawn  more  than  two  equal  sects  to  a 
circle. 

351.  Corollary  I.  If  two  circles  have  three  points  in 
common,  they  coincide. 

Because  from  the  center  of  one  circle  three  equal  sects  can  be  drawn 
to  points  on  the  other  circle, 

/.    they  are  concentric,  and  they  have  a  point  m  common, 

/.     they  coincide. 

(340.  Two  concentric  circles  with  a  point  in  common  coincide.) 


126  THE  ELEMENTS  OF  GEOMETRY. 

352.  Corollary  II.     Through  three  points  not  more  than 
one  circle  can  pass. 

353.  Corollary  III.     Two  different  circles   cannot   meet 
one  another  in  more  than  two  points. 

354.  A   circle   is   circumscribed  about  a  polygon  when   it 
passes  through  all  the  vertices  of  the  polygon. 

Then  the  polygon  is  said  to  be  inscribed  in  the  circle. 


Problem   I. 

355'   Through  any  three  points  not  in  the  same  line^  to  de- 
scribe a  circle. 


yo 


^-r"'/       \2> 


^>x. 


c 


Given,  ihree  points,  A,  B,  and  C,  not  in  t/ie  same  fine. 

Required,  to  describe  a   circle  which  shall  pass   through   all   of 
them. 

Construction.    By  188,  find  the  point  6>,  such  that 

OA  =  OB  =  OC, 

356.  Corollary.     To  circumscribe  a  circle  about  any  given 
triangle,  by  355,  pass  a  circle  through  its  three  vertices. 

357.  Four  or  more  points  which  lie  on  the  same  circle  are 
called  Concyclic, 


THE  CIRCLE.  1 27 


Theorem   XI. 

358.  Perpendiculars  from  the  center  on  equal  chords  are 
equal ;  andy  on  unequal  chords ^  that  on  the  greater  is  the 
lesser. 

C 


Hypothesis.    From  center  O, 

OH,  OK,  OL,  J.  chords  AB  =  CD>  FG, 

Conclusion.     OH  =  OK  <  OL. 

Proof.     Draw  the  radii  OA,  OB,  OC,  OD,  OF. 

By  hypothesis,  AB  =  CD, 

,\     A  OAB  ^  A  OCjD, 
(129.  Triangles  with  three  sides  respectively  equal  are  congruent.) 

.*.     the  altitude  OH  =  corresponding  altitude  OK. 
Again,  because  CZ>  >  FG, 

.'.      CK  >  FL, 

But  'CK'  +  KD*  =  ~0C'  =  OF"  =  FL'  +  T&. 
But  CX'  >  'FL\  _ 

.-.    KO^KLCf", 

/.     OK  <   OL, 

359*  By  33,  Rule  of  Inversion, 

Chords  having  equal  perpendiculars  from  the  center  are 
equal ;  and,  of  chords  having  unequal  perpendiculars,  the  one 
with  the  lesser  is  the  greater. 


128 


THE  ELEMENTS  OF  GEOMETRY. 


II.   Angles   at  the   Center. 

360.  The  explemental  angles  at  the  center  of  a  circle,  whose 
arms  are  the  same  radii,  are  said  to  be  subtended  by^  or  to  stand 
upon^  the  explemental  arcs  opposite  them  intercepted  by  the 
radii,  the  reflex  angle  upon  the  major  arc. 


361.  A  Sector  is  the  figure  formed  by  two  radii  and  the  arc 
included  between  them. 

362.  The  angle  of  the  sector  is  the 
angle  at  the  center  which  stands  upon  the 
arc  of  the  sector. 

363.  A  given  sect  is  said  to  subtend  a 
certain  angle  from  a  given  point  when  the 
lines  drawn  from  the  point  to  the  ends  of  the  sect  form  that 
angle. 

364.  An  inscribed  angle  is  formed  by  two  chords  from  the 
same  point  on  the  circle,  and  is  said  to  stand  upon  the  arc 
between  its  arms. 


j<5t 


THE   CIRCLE.  1 29 


Theorem   XII. 

365.  In  the  same  or  equal  circles ^  equal  arcs  are  congruent^ 
subtend  equal  angles  at  the  center^  determine  equal  sectors^  and 
are  stibtended  by  equal  chords. 


Hypothesis.     Radius  OA  —  radius  CF,  and  arc  AB  =  arc  FG. 

Conclusions.        I.  ^  AOB  =  4.  FCG. 

II.  Sector  AOB  =  sector  FCG. 
III.  Chord  AB  =  chord  FG. 
Proof.     Place  the  sector  A  OB  over  the  sector  FCG  so  that  the 
center  O  shall  fall  on  C,  and  the  radius  OA  on  the  line  CF.    Then, 
because  OA  =  CF^ 

.*.     point  A  falls  on  point  F. 

Again,  because  the  radii  are  equal,  every  point  of  the  arc  AB  will 
fall  on  some  part  of  the  circle  FG. 
But  arc  AB  =  arc  FG, 

.'.    point  B  falls  on  G, 
.'.     chord  AB  =  chord  FG, 
i.AOB  ^  4  FCG, 
and  sector  AOB  =  sector  FCG. 

366.  In  the  same  or  equal  circles  the  stim  of  two  minor  arcs 
is  the  arc  obtained  by  placing  them  on  the  same  circle  so  as 
not  to  overlap,  with  one  end  point  in  common. 


I30  THE  ELEMENTS  OF  GEOMETRY, 


Theorem   XIII. 

367.  A  sum  of  two  arcs  of  the  same  or  equal  circles  subtends 
an  angle  at  the  center  equal  to  the  sum  of  the  angles  which  each 
arc  subtends  separately. 

o 


Proof.  Placing  the  arcs  with  two  end  points  in  common  at  B,  join 
B  and  the  other  end  points,  A  and  C,  to  the  center  O, 

Then  the  angles  are  in  such  a  position,  that,  by  definition  61, 

^  AOB  +  -4.  BOC  =  i.  AOC. 

But  4  AOC  is  subtended  by  arc  AC,  which  is  the  sum  of  arc  AB 
subtending  ^  AOB,  and  arc  BC  subtending  ^  BOC. 

368.  Corollary.  Two  sectors  of  the  same  or  equal  circles 
may  be  so  placed  as  to  form  a  sector  whose  arc  is  the  sum  of 
their  arcs,  whose  angle  is  the  sum  of  their  angles,  and  whose 
surface  is  the  sum  of  their  surfaces. 


Theorem   XIV. 

369.  In  the  same  or  equal  circles,  of  two  uneqtial  arcs,  ths 
greater  subtends  the  greater  angle  at  the  center,  and  determines 
the  greater  sector. 

Proof.  If  the  first  arc  is  greater  than  the  second,  it  is  equal  to  the 
second  plus  a  third  arc;  and  so,  by  367,  the  angle  which  the  first  sub- 
tends is  greater  than  the  angle  which  the  second  subtends  by  the  angle 
which  the  third  arc  subtends  at  the  center. 

And  like  is  true  of  the  sectors. 

370.  From  365  and  369,  by  33,  Rule  of  Inversion, 


THE  CIRCLE.  I31 


In  the  same  or  equal  circles,  equal  angles  at  the  center 
intercept  equal  arcs,  and  determine  equal  sectors ;  and,  of  two 
unequal  angles  at  the  center,  the  greater  intercepts  the  greater 
arc,  and  determines  the  greater  sector. 

371.  Corollary.  A  diameter  of  a  circle  divides  the  en- 
closed surface  into  two  equal  parts. 

372.  Again,  from  365  and  369,  by  33,  Rule  of  Inversion, 

In  the  same  or  equal  circles,  equal  sectors  have  equal  arcs 
and  equal  angles ;  and,  of  two  unequal  sectors,  the  greater  has 
the  greater  arc  and  the  greater  angle. 

Theorem  XV. 

373'  I''^  t^e  same  or  equal  circles y  of  two  unequal  minor  arcSy 
the  greater  is  subtended  by  the  greater  chord ;  of  two  unequal 
major  arcSy  the  greater  is  subtended  by  the  lesser  chord. 

J9. 


D 
Hypothesis.    Minor  arc  AB  >  arc  CD, 

Conclusion.     Chord  AB  >  chord  C£>. 
Proof.     4.  AOB  >  ^  COD, 

(369.  The  greater  arc  subtends  the  greater  angle  at  the  center.) 
.*.     in  As  A  OB  and  COD,  side  AB  >  side  CD. 

(159.  Two  triangles  with  two  sides  equal,  but  the  included  angle  greater  in  the  first, 
have  the  third  side  greater  in  the  first.) 

Secondly,  because  the  minor  arc  with  its  major  arc  together  make 
up  the  entire  circle,  therefore  to  a  greater  major  arc  will  correspond  a 
lesser  minor  arc,  and  therefore  a  lesser  chord. 


132 


THE  ELEMENTS  OF  GEOMETRY. 


374.  From  365  and  373,  by  33,  Rule  of  Inversion, 
In  the  same  or  equal  circles,  equal  chords  subtend  equal 
major  and  minor  arcs;  and,  of  two  unequal  chords,  the  greater 
subtends  the  greater  minor  arc  and  the  lesser  major  arc. 


Theorem   XVI. 

375.  An  angle  at  the  center  of  a  circle  is  double  the  inscribed 
angle  standing  up07t  the  same  arc. 


Fig.  I. 


Fig.  2. 


Hypothesis.  Lei  AB  be  any  arc,  O  the  center,  C  any  point 
on  the  circle  not  on  arc  AB. 

Conclusion.     ^  A  OB  =  2^  A  CB. 
Proof.     Join  CO,  and  produce  it  to  Z>. 
Because,  being  radii,  OA  =  OC, 

.-.  ^  OCA  =  ^  OAC, 
(126.  The  angles  at  the  base  of  an  isosceles  triangle  are  equal.) 

.-.     ^  AOD  =  4.  OCA  4-  4.  OAC  =  2^  OCA. 

(173.  The  exterior  angle  of  a  triangle  equals  the  sum  of  the  interior  opposite 

angles.) 

Similarly,  ^  DOB  =  2^  OCB. 

Hence  (in  Fig.  i)  the  sum  or  (in  Fig.  2)  the  difference  of  the 
angles  AOD,  DOB,  is  double  the  sum  or  difference  of  OCA  and 
OCB\  that  is,  ^  AOB  =  2^  ACB. 


THE   CIRCLE. 


133 


III.  Angles   in   Segments. 

376.  An  angle  made  by  two  lines  drawn  from  a  point  in  the 
arc  of  a  segment  to  the  extremities  of  the  chord  is  said  to  be 
inscribed  in  the  segment,  and  is  called  the  angle  in  the  seg- 


ment. 


377.  Corollary  I.     Angles  inscribed  in  the  same  segment 
of  a  circle  are  equal. 


For  each  of  the  angles  ACBy  ADB  is  half  of  the  angle 
subtended  at  the  center  by  the  arc  ARB, 

378.  Corollary  II.  If  a  circle  is  divided  into  two  seg- 
ments by  a  chord,  any  pair  of  angles,  one  in  each  segment,  will 
be  supplemental. 

For  they  are  halves  of  the  explemental  angles  at  the  center 
standing  on  the  same  explemental  arcs. 


134 


THE  ELEMENTS  OF  GEOMETRY. 


379.  Corollary  III.     The  opposite  angles  of  every  quad- 
rilateral inscribed  in  a  circle  are  supplemental. 


For  they  stand  on  explemental  arcs,  and  so  are  halves  of 
explemental  angles  at  the  center. 

Theorem  XVII. 

-380.  From  a  point  on  the  side  toward  the  segment ^  its  chord 
subtends  an  angle  less  than^  equal  tOy  or  greater  than,  the  angle 
in  the  segment^  according  as  the  point  is  without^  on^  or  within 
the  arc  of  the  segment. 


Hypothesis.  AC  the  chord  of  any  segment,  ABC;  P  any  point 
without  the  segment  on  the  same  side  of  AC  as  B ;  Q  any  point 
within  the  segment. 

Conclusions.  Since,  by  377,  we  know  all  angles  inscribed  in  the 
segment  are  equal,  we  have  only  to  prove 

I.  ^  AFC  <  4.  ABC. 
II.  4AQC>  4  ABC, 


THE   CIRCLE.  1 35 


Proof.     I.  Let  R  be  the  point  where  PC  meets  the  circle,  and  join 
RA,  making  A  APR)  then  4.  APC  <4.ARC, 
(142.  An  exterior  angle  of  a  triangle  is  greater  than  either  interior  opposite  angle.) 

II.  For  the  same  reason,  if  ^^  is  produced  to  meet  the  circumfer- 
'ence  at  S,  and  SC  joined,  making  A  CSQ,  4-  ^Q.^ >  4-  CSA. 

381.  By  33,  Rule  of  Inversion, 

On  the  side  toward  a  segment,  the  vertex  of  a  triangle,  with 
its  chord  as  base,  will  lie  without,  on,  or  within  the  arc  of  the 
segment,  according  as  the  vertical  angle  is  less  than,  equal  to, 
or  greater  than,  an  angle  in  the  segment. 


Theorem  XVIII. 

382.  If  two  opposite  angles  of  a  quadrilateral  are  supple- 
mental, a  circle  passing  through  any  three  of  its  vertices  will 
contaifz  the  fourth. 


Hypothesis.    ABCD  is  a  quadrilateral,  with  ^  A  -\-  ^  C  =  st.  ^. 

Conclusion.  The  four  vertices  lie  on  the  circle  determined  by  any 
three  of  them. 

Proof.     By  355,  pass  a  circle  through  the  three  points  B,  C,  D. 

Take  any  point,  F,  on  the  arc  DFB  of  the  segment,  on  the  same 
side  of  DB  as  A.    Join  FB,FD.    Then^  7^  +  ^  C  =  st.  ^. 

{379.  The  opposite  angles  of  an  inscribed  quadrilateral  are  supplemental.) 

From  hypothesis,  .*.     ^  A  =  :4  F,         .*.     ^  is  on  the  arc  DFB. 

(381.  On  the  side  toward  a  segment,  the  vertex  of  a  triangle,  with  its  chord  as  base, 
will  lie  on  its  arc  if  the  vertical  angle  is  equal  to  an  angle  in  the  segment.) 


136  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XIX. 

383.  The  angle  in  a  segment  is  greater  than^  equal  to^  or  less 
than,  a  right  angle,  accordijtg  as  the  arc  of  the  segment  is  less 
than,  equal  to,  or  greater  than,  a  semicircle. 


Hypothesis.  Lei  AD  be  a  diameter  of  a  circle  whose  center 
is  O.     Take  B  and  C  points  on  the  same  circle. 

Conclusions.     I.  ^  ACD  =  rt.  2^,  being  half  the  straight  angle 

A  on. 

(375.  An  angle  at  the  center  is  double  the  inscribed  angle  on  the  same  arc.) 

II.     .-.     ^ADC<xt.4.. 
(143.  Any  two  angles  of  a  triangle  are  together  less  than  a  straight  angle.) 

III.     .-.     4.ABC>x\..4., 
since 

4.ADC  -\-  }^ABC  =  st.  4.. 

(379.  Opposite  angles  of  an  inscribed  quadrilateral  are  supplemental.) 

384.  By  33,  Rule  of  Inversion, 

A  segment  is  less  than,  equal  to,  or  greater  than,  a  semi- 
circle, according  as  the  angle  in  it  is  greater  than,  equal  to,  or 
less  than,  a  right  angle. 


THE   CIRCLE.  1 37 


Theorem  XX. 

385.  If  two  chords  intersect  within  a  circle y  an  angle  formed 
and  its  vertical  are  each  equal  to  half  the  angle  at  the  center 
standing  on  the  sum  of  the  arcs  they  intercept. 


Hypothesis.    Let  the  chords  AC,BD  intersect  at  F  within  the 
circle. 

Conclusion.     ^  BFC  =  half  ^  at  center  standing  on  (arc  BC  + 
arc  DA) . 

Proof.    Join  CB>. 

4.  BFC  =  %.  BDC  +  4.  DC  A, 

(173.  The  exterior  angle  of  a  triangle  is  equal  to  the  sum  of  the  opposite  interior 

angles.) 

But  2  4.  BDC  =  ^  at  center  on  BC,  and  2  4  DCA  =  4  at  cen- 
ter on  DA ; 

(375.  An  angle  at  the  center  is  double  the  inscribed  angle  upon  the  same  arc.) 

.-.     2  4  BFC  —  4a.t  center  on  arc  (BC  +  DA). 

(367.  A  sum  of  two  arcs  subtends  an  angle  at  the  center  equal  to  the  sum  of  the 
angles  subtended  by  the  arcs.) 

Exercises.     71.  The  end  points  of  two  equal  chords  of  a 
circle  are  the  vertices  of  a  symmetrical  trapezoid. 

72.  Every  trapezoid  inscribed  in  a  circle  is  symmetrical. 


138  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XXI. 

386.  An  angle  formed  by  two  secants  is  half  the  angle  at 
center  standing  on  the  difference  of  the  arcs  they  intercept. 


Hypothesis.    Lei  two  lines  from  F  cut  the  circle  whose  center 
is  O,  in  the  points  A,  B,  C,  and  D. 

Conclusion.     ^  ^  =  |  2j^  at  (9  on  difference  between  arc  AB  and 
arc  CD. 

Proof.    Join^C 

^  CAD  =  4.F^-  4.C. 

Doubling  both  sides,  :^  at  6>  on  arc  CD  =2^i^-f^at^0n 
arc  AB ; 

.*.     twice  ^  F  =  difference  of  2j^  s  at  O  on  arcs  CD  and  AB. 


IV.   Tangents. 

387.  A  line  which  will  meet  the  circle  in  one  point  only 
is  said  to  be  a  Tangent  to  the  circle. 

388.  The  point  at  which  a  tangent  touches  the  circle  is 
called  the  Point  of  Contact. 


THE   CIRCLE.  1 39 


Theorem   XXII. 

389.  Of  lines  passiiig  through  the  e7id  of  any  radius,  the 
perpendicular  is  a  tangent  to  the  circley  and  every  other  line  is  a 
secant. 


Hypothesis.    A  radius  OP  perpendicular  fo  PB,  oblique  to  PC. 

Conclusions.      I.  PB  is  a  tangent  at  P. 

II.  PC  is  a  secant. 
Proof.     (I.)    The  sect  from   O  to  any  point  on  PB,  except  P, 
is  >  OP, 

(150.  The  perpendicular  is  the  least  sect  between  a  point  and  a  line.) 

.*.     every  point  of  PB  except  P  is  outside  the  circle. 

{321.  A  point  is  without  a  circle  if  its  sect  from  the  center  is  greater  than  the 

radius.) 

Proof.     (II.)  A  sect  perpendicular  to  PC  is  less  than  the  oblique 

OP, 

.*.     a  point  of  PC  is  within  the  circle ; 

/.     PC  is  a  secant. 

390.  Corollary  I.  One  and  only  one  tangent  can  be 
drawn  to  a  circle  at  a  given  point  on  the  circle. 

391.  Corollary  II.  To  draw  a  tangent  to  a  circle  at  a 
point  on  the  circle,  draw  the  perpendicular  to  the  radius  at  the 
point. 


140 


THE  ELEMENTS  OF  GEOMETRY. 


392.  Corollary  III.  The  radius  to  the  point  of  contact 
of  any  tangent  is  perpendicular  to  the  tangent. 

393*  Corollary  IV.  The  perpendicular  to  a  tangent  from 
the  point  of  tangency  passes  through  the  center  of  the  circle. 

394.  Corollary  V.  The  perpendicular  drawn  from  the 
center  to  the  tangent  passes  through  the  point  of  contact. 

On  the  Three  Relative  Positions  of  a  Line  and  a 

Circle. 

395'  Corollary  VI.  A  line  will  be  a  secant,  a  tangent,  or 
not  meet  the  circle,  according  as  its  perpendicular  from  the 
center  is  less  than,  equal  to,  or  greater  than,  the  radius. 

396.  By  33,  Rule  of  Inversion, 

The  perpendicular  on  a  line  from  the  center  will  be  less 
than,  equal  to,  or  greater  than,  the  radius,  according  as  it  is  a 
secant,  tangent,  or  non-meeter. 


Theorem  XXIII. 

397.  An  angle  formed  by  a  tangent  and  a  chord  from  the 
point  of  contact  is  half  the  angle  at  the  center  standing  on  the 
intercepted  arc.  y 


Htpothesis,     AB  is  tangeni  at  C,  and  CD  is  a  chord  of  O 
wUh  center  at  O. 


THE   CIRCLE.  141 


Conclusions.     ^DCB  =  ^^DOC, 

4.DCA  =  |explementZ>(9C 

Proof.     At  C  erect  chord  CF  Jl  AB. 

CF  is  a  diameter  of  the  circle. 

(393.  The  perpendicular  to  a  tangent  from  the  point  of  contact  passes  through  the 
center  of  the  circle.) 

Join  OD.    Then 

rt.  ^  OCA  =  l^st.  ^at  O, 
also 

^  OCD  =  i^  FOB, 

(375.  The  angle  at  the  center  is  double  the  inscribed  angle  on  the  same  arc.) 

Therefore,  adding 

^  £>CA  =  ^reflex  ^  Z>OC, 

therefore  the  supplement  of  ^  VGA,  which  is  ^  DCB,  is  half  the  exple- 
ment  of  reflex  ^  DOC,  which  is  4.  DOC, 

398.  Inverse.  If  the  angle  at  the  center  standing  on  the 
arc  intercepted  by  a  chord  equals  twice  the  angle  made  by  that 
chord  and  a  line  from  its  extremity  on  the  same  side  as  the 
arc,  this  line  is  a  tangent. 

Proof.  There  is  but  one  line  which  will  make  this  angle ;  and  we 
already  know,  from  397,  that  a  tangent  makes  it. 

Exercises.  73.  The  chord  which  joins  the  points  of  con- 
tact of  parallel  tangents  to  a  circle  is  a  diameter. 

74.  How  may  397  be  considered  as  a  special  case  of  375  1 

75.  A  parallelogram  inscribed  in  a  circle  must  be  a  rect- 
angle. 

y6.  If  a  series  of  circles  touch  a  given  line  at  a  given  point, 
where  will  their  centers  all  lie } 

'jj.  The  angle  of  two  tangents  is  double  that  of  the  chord 
of  contact  and  the  diameter  through  either  point  of  contact. 


142  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XXIV. 

399.  The  angle  formed  by  a  tangent  and  a  secant  is  half  the 
angle  at  the  center  standing  on  the  difference  of  the  intercepted 
arcs. 


Hypothesis.  Of  iwo  lines  from  Z>,  one  cuts  at  B  and  C  the  O 
whose  center  is  O,  the  other  is  tangent  to  the  same  O  at  A. 

Conclusion.     2  ^  D  —  difference  between  ^  AOC  and  ^  AOB. 
Proof.    Join  AB. 

^  ABC  =  ^D  +  ^  DAB, 
(173.  The  exterior  angle  of  a  triangle  is  equal  to  the  two  interior  opposite  angles.) 

.-.     4.  AOC  =  24.D  +  :^  AOB, 

(375.  An  angle  at  the  center  is  double  the  inscribed  angle  on  the  same  arc.) 

and 

(397.  An  angle  formed  by  a  tangent  and  chord  is  half  the  angle  at  the  center  on 

the  intercepted  arc.) 

.-.     twice  ^  Dis  the  difference  between  ^  AOC  and  ^  AOB. 

400.  Corollary.  The  angle  formed  by  two  tangents  is 
half  the  angle  at  the  center  standing  on  the  difference  of  the 
intercepted  arcs. 


THE   CIRCLE.  I43 


Problem   II. 

401.  From  a  given  point  without  a  circle  to  draw  a  tangent 
to  the  circle. 


Given,  a  O  ly/M  center  O,  and  a  point  P  outside  it 

Required,  to  draw  through  P  a  tangent  to  the  circle. 
Construction.    Join  OP.    Bisect  OP  in  C. 
With  center  C  and  radius  CP  describe  a  circle  cutting  the  given 
circle  in  F  and  G. 

Join  PF  and  PG. 

These  are  tangents  to  O  FGB. 
Proof.    Join  OF. 

^  OFF  is  a  rt.  ^, 
(383.  The  angle  in  a  semicircle  is  a  right  angle.) 

.-.    PF  is  tangent  at  i^  to  O  BFG. 
(389.  A  line  perpendicular  to  a  radius  at  its  extremity  is  a  tangent  to  the  circle.) 

402.  Corollary.  Two  tangents  drawn  to  a  circle  from  the 
same  external  point  are  equal,  and  make  equal  angles  with 
the  line  joining  that  point  to  the  center. 


144 


THE  ELEMENTS  OF  GEOMETRY. 


V.  Two   Circles. 

Theorem   XXV. 

403.  The  line  joining  the  centers  of  two  circles  which  meet 
in  two  points  is  identical  with  the  perpendicular  bisector  of  the 
common  chord. 


Proof.  For  the  perpendicular  bisector  of  the  common  chord  must 
pass  through  the  centers  of  the  two  circles. 

(331.  The  perpendicular  bisector  of  any  chord  passes  through  the  center.) 

404.  Corollary.  Since  any  common  chord  is  bisected  by 
the  line  joining  the  centers,  therefore  if  the  two  circles  meet 
at  a  point  on  the  line  of  centers,  there  is  no  common  chord, 
and  these  circles  have  no  second  point  in  common. 

405.  Two  circles  which  meet  in  one  point  only  are  said  to 
touch  each  other,  or  to  be  tangent  to  one  another,  and  the  point 
at  which  they  meet  is  called  their  point  of  contact. 

406.  By  343, 

Two  circles^  not  concentric^  have  always  one,  and  only  one^ 
common  axis  of  symmetry ;  namely,  their  line  of  centers. 

For  this  is  the  only  line  which  contains  a  diameter  of 
each. 


THE   CIRCLE.  1 45 


Theorem   XXVI. 


Obverse  of  404. 


407*  Jf  t'^0  circles  have  one  common  point  not  on  the  line 
through  their  centers^  they  have  also  another  common  point. 


Hypothesis.  O  with  center  O  and  O  with  center  C,  having  a 
common  point  B  not  on  OC. 

Conclusion.    They  have  another  common  point. 
Proof.    Join  OC,  and  from  B  drop  a  line  perpendicular  to  OC  2X 
D,  and  prolong  it,  making  DF  =  BD. 

F  is  the  second  common  point. 
For        A  ODB  ^  A  OFD,        and        A  CBD  ^  A  CFD-, 
(124.  Triangles  having  two  sides  and  the  included  angle  equal  in  each  are  con- 
gruent.) 

.-.     OF  =  OB,        and         CF  =^  CB -, 

.'.    F  is  on  both  Os. 

{321.  A  point  is  on  the  circle  if  its  distance  from  the  center  is  equal  to  the  radius.) 

408.  CONTRANOMINAL    OF    40/.       If    tWO    circIcS    tOUCh    OHC 

another,  the  line  through  their  centers  passes  through  the  point 
of  contact. 

409.  Corollary.  Two  circles  which  touch  one  another 
have  a  common  tangent  at  their  point  of  contact ;  namely,  the 
perpendicular  through  that  point  to  the  line  joining  their 
centers. 


146 


THE  ELEMENTS  OF  GEOMETRY, 


410.  Calling  the  sect  joining  the  centers  of  two  circles  their 
center-sect  c,  and  calling  their  radii  r^  and  rj,  we  have,  in 
regard  to  the  relative  positions  of  two  circles,  — 

I.  If  ^>ri  +  ^2,  therefore  the  Os  are  wholly  exterior. 


2.  If  ^  =  /'j  +  r^y  therefore  the  Os  touch  externally. 


3.  \i  c  <r^-\-  ^2,  but  c  >  the  difference  of  radii,  therefore 
the  Os  cut  each  other. 


THE   CIRCLE.  1 47 


4.  If  ^  =  the  difference  of  radii,  therefore  the  Os  touch 
internally. 


5.  If  ^  <  the  difference  of  radii,  therefore  one  0  is  wholly 
interior  to  the  other. 


411.  By  33,  Rule  of  Inversion,  the  five  inverses  to  the 
above  are  true. 

Exercises.  78.  How  must  a  line  through  one  of  the  com- 
mon points  of  two  intersecting  circles  be  drawn  in  order  that 
the  two  circles  may  intercept  equal  chords  on  it  ? 

79.  Through  one  of  the  points  of  intersection  of  two  circles 
draw  the  line  on  which  the  two  circles  intercept  the  greatest 
sect. 

80.  If  any  two  lines  be  drawn  through  the  point  of  contact 
of  two  circles,  the  lines  joining  their  second  intersections  with 
each  circle  will  be  parallel. 


148 


THE  ELEMENTS  OF  GEOMETRY, 


VI.   Problems. 


Problem   III. 
412.  To  bisect  a  given  arc. 

Given,  the  arc  BD. 
Required,  to  bisect  it. 

Construction.    Join  BD,  and  bisect  the  sect  BD 
m  F;  2X  F  erect  a  perpendicular  cutting  the  arc  in  C. 

C  is  the  mid  point  of  the  arc. 

Proof.    Join  BC,  CD. 

aBCF^  aDFC, 

(124.  Triangles  having  two  sides  and  the  included  angle  in  each  equal  are  con- 
gruent.) 

.-,     chord  BC  =  chord  CD, 
a.TcBC  =  arc  CD. 
(374.  In  the  same  circle,  equal  chords  subtend  equal  minor  arcs.) 

413.  A  polygon  is  said  to  be  circumscribed  about  a  circle 
when  all  its  sides  are  tangents  to  the  circle. 


The  circle  is  then  said  to  be  inscribed  in  the  polygon. 
414.  A  circle  which  touches  one  side  of  a  triangle  and  the 
other  two  sides  produced  is  called  an  Escribed  Circle. 


THE   CIRCLE. 


149 


Problem   IV. 

415.   To  describe  a  circle  touching  three  given  lines  which  are 
not  all  parallel^  and  do  not  all  pass  through  the  same  point. 


Given,  three  lines  interseciing  in  A^  B,  and  C, 

Required,  to  describe  a  circle  touching  them. 

Construction.     Draw  the  bisectors  of  the  angles  at  A  and  C. 

These  four  bisectors  will  intersect  in  four  points,  (9,  (9i,  O^,  Oy 

A  circle  described  with  any  one  of  these  points  as  center,  and  its 
perpendicular  on  any  one  of  the  three  given  lines  as  radius,  will  touch 
all  three. 

Proof.  By  186,  every  point  in  the  bisector  of  an  angle  is  equally 
distant  from  its  arms  ; 

Therefore,  since  O  is  on  the  bisector  of  ^  A,  the  perpendicular 
from  O  on  AB  equals  the  perpendicular  from  O  on  AC,  which  also 
equals  the  perpendicular  from  O  on  CB,  since  O  is  also  on  the  bisector 

of  ^  a 


150  THE  ELEMENTS  OF  GEOMETRY. 

416.  The  four  tangents  common  to  two  circles  occur  in  two 
pairs  intersecting  on  the  common  axis  of  symmetry. 


Problem  V. 

417.  In  a  given  circle  to  inscribe  a  triangle  equiangular  to  a 
given  triangle. 


Given,  a  0  and  A  ABC. 

Required,  to  describe  in  the  (d  a  t^  equiangular  to  A  ABC. 
Construction.     Draw  a  tangent  GH  touching  the   circle   at  the 
point  D. 

Make  4.  HDK  =  2^:  C,  and  4.  GDL  =  ^  A. 
K  and  L  being  on  the  circumference,  join  KL. 
DKL  is  the  required  triangle. 
Proof.     ^  KDL  =  ^  B. 

(174.  The  three  angles  of  a  triangle  are  equal  to  a  straight  angle.) 

4K  =  7^A,         and         ^  L  =  }^  C. 

(375'  An  inscribed  angle  is  half  the  angle  at  the  center  on  the  intercepted  arc.) 

(397.  An  angle  formed  by  a  tangent  and  a  chord  is  half  the  angle  at  the  center 

on  the  intercepted  arc.) 


BOOK    IV. 


REGULAR  POLYGONS. 
I.   Partition   of  a  Perigon. 

Problem   I. 
418.   To  bisect  a  perigon. 


Solution.  To  bisect  the  perigon  at  the  point  (9,  draw  any  line 
through  O. 

This  divides  the  perigon  into  two  straight  angles,  and  all  straight 
angles  are  equal. 

419.  Corollary.  By  drawing  a  second  line  through  O  at 
right  angles  to  the  first,  we  cut  the  perigon  into  four  equal 
parts ;  and  as  we  can  bisect  any  angle,  so  we  can  cut  the  peri- 
gon into  8,  16,  32,  64,  etc.,  equal  parts. 


152  THE  ELEMENTS  OF  GEOMETRY. 

Problem   II. 
420.  To  trisect  a  perigon. 


Solution.  To  trisect  the  perigon  at  the  point  O,  to  O  draw  any 
line  BO;  on  BO  produced  take  a  sect  0C\  on  OC  construct,  by  132, 
an  equilateral  triangle  CDO. 

^  DOB  is  one-third  of  a  perigon. 

For  ^  DOC  i?,  one-third  of  a  st.  :^, 

(174.  The  three  angles  of  a  triangle  are  equal  to  a  straight  angle.) 
.*.     ^  DOB  is  two-thirds  of  a  st.  2(^. 

421.  Corollary.  Since  we  can  bisect  any  angle,  so  we 
may  cut  the  perigon  into  6,  12,  24,  48,  etc.,  equal  parts. 

422.  Remark.  To  trisect  a7ty  given  angle  is  a  problem 
beyond  the  power  of  strict  Elementary  Geometry,  which  allows 
the  use  of  only  the  compasses  and  an  unmarked  ruler.  There 
is  an  easy  solution  of  it,  which  oversteps  these  limits  only  by 
using  two  marks  on  the  straight-edge.  The  trisection  of  the 
angle,  the  duplication  of  the  cube,  and  the  quadrature  of  the 
circle,  are  the  three  famous  problems  of  antiquity. 


REGULAR  POLYGONS:  1 53 


Problem   III. 
423.   To  cut  a  perigon  mto  five  equal  parts. 


Solution.    By  314,  describe  an  isosceles  triangle  -^4-5C  having 

^A  =  4.C^24.B, 
Then  ^  ^  is  two-fifths  of  a  st.  ^ . 

(174.  The  three  angles  of  any  triangle  are  equal  to  a  straight  angle.) 

Therefore,  to  get  a  fifth  of  a  perigon  at  a  point  O,  construct,  by  164, 
4.  GOB  =  ^A. 

424.  Corollary.     Since  we  can  bisect  any  angle,  we  may- 
cut  a  perigon  into  10,  20,  40,  80,  etc.,  equal  parts. 

Problem   IV. 

425.  To  cut  a  perigon  into  fifteen  equal  parts, 

'J3 


Solution.  At  the  perigon  point  (9,  by  420,  construct  the  ^  AOC 
=  one-third  of  a  perigon. 

By  423,  make  2^  AOB  =  one-fifth  of  a  perigon. 

Then  of  such  parts,  as  a  perigon  contains  fifteen,  -i^  AOC  contains 
five,  and  ^  AOB  contains  three,  therefore  ^  BOC  contains  two. 

So  bisecting  ^  BOC  giwts  one-fifteenth  of  a  perigon. 


154 


THE  ELEMENTS  OF  GEOMETRY, 


426.  Corollary.     Hence  a  perigon  may  be  divided  into  30, 
60,  120,  etc.,  equal  parts. 


II.   Regular   Polygons   and   Circles. 

Problem   V. 

427.   To  mscribe  i?t  a  circle  a  regular  polygon  having  a  given 
number  of  sides.  c 


This  problem  can  be  solved  if  a  perigon  can  be  divided  into  the 
given  number  of  equal  parts. 

For  let  the  perigon  at  O,  the  center  of  the  circle,  be  divided  into  a 
number  of  equal  parts,  and  extend  their  arms  to  meet  the  circle  in  Ay 
By  C,  D,  etc.     Draw  the  chords,  AB,  BC,  CD,  etc. 

Then  shall  ABCD,  etc.,  be  a  regular  polygon. 

For  if  the  figure  be  turned  about  its  center  O,  until  OA  coincides 
with  the  trace  of  OB,  therefore,  because  the  angles  are  all  equal,  OB 
will  coincide  with  the  trace  of  OC,  and  6>C  with  the  trace  of  OD,  etc. ; 
then  AB  will  coincide  with  the  trace  of  BC,  and  BC  with  the  trace  of 
CD,  etc. ; 

.-.    AB  =  BC  ==  CD  =  etc., 

.•4     the  polygon  is  equilateral. 
Moreover,  since  then  ABC  will  coincide  with  the  trace  of  BDCy 
.-.     t  ^BC  =  4.  BCD  =  etc., 
.*.     the  polygon  is  equiangular. 
Therefore  ABCD,  etc.,  is  a  regular  polygon,  and  it  is  inscribed  in 
the  given  circle. 


REGULAR  POLYGONS.  1 55 

428.  Remark.  From  the  time  of  Euclid,  about  300  B.C., 
no  advance  was  made  in  the  inscription  of  regular  polygons 
until  Gauss,  in  1796,  found  that  a  regular  polygon  of  17  sides 
was  inscriptible,  and  in  his  abstruse  Arithmetic,  published  in 
1 80 1,  gave  the  following  :  — 

In  order  that  the  geometric  division  of  the  circle  into  n 
parts  may  be  possible,  n  must  be  2,  or  a  higher  power  of  2,  or 
else  a  prime  number  of  the  form  2"^  +  i,  or  a  product  of  two 
or  more  different  prime  numbers  of  that  form,  or  else  the 
product  of  a  power  of  2  by  one  or  more  different  prime  num- 
bers of  that  form. 

In  other  words,  it  is  necessary  that  n  should  contain  no  odd 
divisor  not  of  the  form  2"'  +  i,  nor  contain  the  same  divisor  of 
that  form  more  than  once. 

Below  300,  the  following  38  are  the  only  possible  values  of 
/^:  2,  3,  4,  5,  6,  8,  10,  12,  15,  16,  17,  20,  24,  30,  32,  34,  40,  48, 
51,  60,  64,  6Z,  80,  85,  96,  102,  120,  128,  136,  160,  170,  192,  204, 
240,  255,  256,  257,  272. 

Exercises.  81.  The  square  inscribed  in  a  circle  is  double 
the  square  on  the  radius,  and  half  the  square  on  the  diame- 
ter. 

^2,  Prove  that  each  diagonal  is  parallel  to  a  side  of  the 
regular  pentagon. 

83.  An  inscribed  equilateral  triangle  is  equivalent  to  half  a 
regular  hexagon  inscribed  in  the  same  circle. 

84.  An  equilateral  triangle  described  on  a  given  sect  is 
equivalent  to  one-sixth  of  a  regular  hexagon  described  on  the 
same  sect. 

85.  If  a  triangle  is  equilateral,  show  that  the  radius  of 
the  circumscribed  circle  is  double  that  of  the  inscribed  ;  and  the 
radius  of  an  escribed,  triple. 

%6.  The  end  points  of  a  sect  slide  on  two  lines  at  right 
angles  :  find  the  locus  of  its  mid-point. 


156 


THE  ELEMENTS  OF  GEOMETRY. 


Problem  VI. 

429.  To  circumscribe  about  a  given  circle  a  regular  polygon 
having  a  given  number  of  sides. 


JW C  X' 


This  problem  can  be  solved  if  a  perigon  can  be  divided  into  the 
given  number  of  equal  parts. 

For  let  the  perigon  at  O,  the  center  of  the  circle,  be  divided  into  a 
number  of  equal  angles,  and  extend  their  arms  to  meet  the  circle  in  A^ 
Bj  C,  D,  etc.     Draw  perpendiculars  to  these  arms  at  A,  B^  C,  D,  etc. 

These  will  be  tangents. 

Call  their  points  of  intersection  K,  L,  M,  etc. 

Then  shall  KLM,  etc.,  be  a  regular  polygon. 

For  if  the  figure  be  turned  about  its  center  O  until  OA  coincides 
with  the  trace  of  OB,  then,  because  the  angles  are  all  equal,  OB  will 
coincide  with  the  trace  of  OC,  and  6>Cwith  the  trace  of  OD,  etc. 

Therefore  the  tangents  at  A,  B,  C,  etc.,  will  coincide  with  the  traces 
of  the  tangents  at  B,  C,  D,  etc. 

Hence  the  polygon  will  coincide  with  its  trace ; 


also 


.-.     KL  =  LM  =  etc., 
^K  =  :^L  =  :^M  =  etc. ; 


therefore  the  polygon  is  regular,  and  it  is  circumscribed  about  the  given 
circle. 


REGULAR  POLYGONS.  1 5/ 

430.  Corollary.  Hence  we  can  circumscribe  about  a  cir- 
cle regular  polygons  of  3,  4,  5,  6,  8,  10,  12,  15,  16,  17,  etc., 
sides. 


Problem  VII. 
431.  To  circumscribe  a  circle  about  a  given  regular  polygon. 


Given,  a  regular  polygon,  as  ABCDE, 

Required,  to  describe  a  circle  about  it. 

Construction.     Bisect  -4.  EAB  and  4-  ABC  by  lines  intersecting  in 
O.    With  center  O  and  radius  OA  describe  a  circle. 
This  shall  be  the  required  circle. 
Proof.    Join  OC.    Then 

A  OBC'^  A  OBAy 

(124.  Triangles  having  two  sides  and  the  included  angle  in  each  equal  are  con- 
gruent.) 

.*.     ^  OCB  =  4  OAB  =  half  one  ^  of  the  regular  polygon, 

.-.     4  B  CI?  is  bisected. 
Similarly  prove  each  ^  of  the  polygon  bisected, 

.-.    OA  =  OB  =  oc  =  on  =  OE, 

(148.  In  a  triangle,  sides  opposite  equal  angles  are  equal.) 

therefore  a  circle  with  radius  OA  passes  through  B^  C,  D,  Ey  and  is 
circumscribed  about  the  given  polygon. 


158  THE  ELEMENTS  OF  GEOMETRY, 

Problem  VIII. 
432.   To  inscribe  a  circle  in  a  given  regular  polygon. 


T 


c 


Given,  a  regular  polygon,  ABODE, 

Required,  to  inscribe  a  circle  in  it. 

Construction.  Bisect  4.  EAB  and  ^  ABC  by  lines  intersecting  in 
O.     From  O  drop  OP  perpendicular  to  AB. 

The  O  with  center  O  and  radius  OP  shall  be  the  O  required. 

Proof.  Join  00,  OD,  OE,  and  draw  0Q1.B0,  OR  A.  CD,  OS 
±  DEy  and  OT  J.  EA,    Then 

A  OBC^  A  OBA, 

(124.  Triangles  having  two  sides  and  the  included  angle  in  each  equal  are  con- 
gruent.) 

.*.     ^  OCB  =  2^  OAB  =  half  one  ^  of  the  regular  polygon, 

.•;     ^  BCD  is  bisected. 

Similarly  prove  each  ^  of  the  polygon  bisected. 
Again, 

A  OBP  ^  A  OBQ, 

(176.  Triangles  having  two  angles  and  a  corresponding  side  in  each  equal  are  con- 
gruent.) 

/.     OP  =  OQ, 


REGULAR  POLYGONS.  1 59 

Similarly, 

OQ  =^  OR  =  OS  =  OT, 

therefore  a  circle  with  radius  OF  will  touch  AB,  BC,  CD,  DE,  EA, 
at  points  P,  Q,  R,  S,  T; 

.'.     it  is  inscribed  in  the  given  polygon. 

433*  Corollary  I.  The  inscribed  and  circumscribed  cir- 
cles of  a  regular  polygon  are  concentric. 

434.  Corollary  II.  The  bisectors  of  the  angles  of  a  reg- 
ular polygon  all  meet  in  a  point  which  is  the  center  both  of  the 
circumscribed  and  inscribed  circles,  and  is  called  the  center 
of  the  regular  polygon. 

435'  Corollary  III.  The  perpendicular  bisectors  of  the 
sides  of  a  regular  polygon  all  pass  through  its  center. 

436.  The  radius  of  its  circumscribed  circle  is  called  the 
radius  of  a  regular  polygon.  The  radius  of  its  inscribed  circle 
is  called  its  apothem. 

437.  The  side  of  a  regular  hexagon  inscribed  in  a  circle  is 
equal  to  the  radius.  For  the  sects  from  the  center  to  the  ends 
of  a  side  make  an  isosceles  triangle,  one  of  whose  angles  is 
one-third  a  straight  angle ;  therefore  it  is  equilateral. 


III.   Least   Perimeter    in    Equivalent    Figures.  —  Greatest 
Surface   in   Isoperimetric   Figures. 

438.  Any  two  figures  are  called  Isoperimetric  when   their 
perimeters  are  equal. 


l60  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   I. 

439'  Of  all  equivalent  triangles  having  the  same  base^  that 
which  is  isosceles  has  the  least  perimeter. 


Hypothesis.    Lei  ABC  be  an  isosceles  iriangle,  and  ABC  any 
equivalent  iriangle  having  ihe  same  base. 

Conclusion.    AB  +  AC<A'B  +  A'C, 
Proof.    A  A'  \\  BC. 

(253.  Inverse.    Equivalent  triangles  on  the  same  base,  and  on  the  same  side  of  it, 
are  between  the  same  parallels.) 

Draw  CND  ±  AA\  meeting  BA  produced  in  D.    Join  A'D, 

4.NAC  =  ^ACB, 
(168.  If  a  transversal  cuts  two  parallels,  the  alternate  angles  are  equal.) 

•4.ACB  =  ^ABC, 
(126.  In  an  isosceles  triangle  the  angles  opposite  the  equal  sides  are  equal.) 

^  ABC  =  4.  DAN, 
(169.  If  a  transversal  cuts  two  parallels,  the  corresponding  angles  are  equal.) 

.-.     i^ACN^  ^ADN, 

(128.  Triangles  having  two  angles  and  the  included  side  equal  in  each  are  con- 
gruent.) 

/.    AN  is  the  perpendicular  bisector  of  CD, 

.-.    AD  =  AC,  and  A'D  =  A' C. 

(183.  The  locus  of  the  point  to  which  sects  from  two  given  points  are  equal,  is  the 
perpendicular  bisector  of  the  sect  joining  them.) 


REGULAR  POLYGONS.  l6l 

But 

BD  <  A'B  +  A'D, 
(156.  Any  two  sides  of  a  triangle  are  together  greater  than  the  third.) 

.-.     AB  ^  AC<A'B  ■\-  AC, 

440.  Corollary.  Of  all  equivalent  triangles,  that  which 
is  equilateral  has  the  least  perimeter. 

For  the  triangle  having  the  least  perimeter  enclosing  a 
given  surface  must  be  isosceles  whichever  side  is  taken  as  the 
base. 

Theorem   II. 

441.  Of  all  isoperimetric  triangles  having  the  same  hase^ 
that  which  is  isosceles  has  the  greatest  surface. 

MAS 


Hypothesis.  Lei  ABC  be  an  isosceles  friangle ;  and  lei  A'BC 
standing  on  ihe  same  base  BC,  have  an  equal  perimeier;  thai  is, 

A'B  +  A'C  =  AB  +  AC. 

Conclusion,     a  ABC  >  a  A'BC. 

Proof.  The  vertex  A'  must  fall  between  BC  and  the  parallel  AN\ 
since,  if  it  fell  upon  AN,  by  the  preceding  proof,  A'B  •\-  A' C  >  AB  + 
AC\  and,  if  it  fell  beyond  AN,  the  sum  A'B  +  A'C  would  be  still 
greater. 

Therefore  the  altitude  of  A  ABC  is  greater  than  the  altitude  of 
A  A'BC,  and  hence  also  its  surface. 

442.  Corollary.  Of  all  isoperimetric  triangles,  that  which 
is  equilateral  is  the  greatest. 

For  the  greatest  triangle  having  a  given  perimeter  must  be 
isosceles  whichever  side  is  taken  as  the  base. 


1 62 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   III. 

443.  Of  all  triangles  formed  with  the  same  two  given  sides ^ 
that  in  which  these  sides  are  perpendicular  to  each  other  has  the 
greatest  surface. 


Hypothesis.     Lei  ABC,  A'BCy  be  two  triangles  having  the  sides 
AB,  BC,  respectively  equal  to  A'B,  BC\  and  let  ^  ABC  be  right 

Conclusion,     a  ABC  >  A  A'BC. 

Proof.     Taking  BC  sls   the   common   base,  the  altitude  AB  of 
A  ABC  is  greater  than  the  altitude  A'B>  of  A  A'BC. 


Theorem   IV. 

444.  Of  all  isoperimetric  plane  figures^  the  circle  contains 
the  greatest  surface. 


Proof.  With  a  given  perimeter,  there  may  be  an  indefinite  number 
of  figures  differing  in  form  and  size.  The  surface  may  be  as  small  as 
we  please,  but  cannot  be  increased  indefinitely. 

Therefore,  among  all  the  figures  of  the  same  perimeter,  there  must 
be  one  greatest  figure,  or  several  equivalent  greatest  figures  of  different 
forms. 


REGULAR  POLYGONS.  1 63 

Every  closed  figure,  if  the  greatest  of  a  given  perimeter,  must  be 
convex;  that  is,  such  that  any  sect  joining  two  points  of  the  perimeter 
Hes  wholly  within  the  figure.  For  let  ACBNA  be  a  non-convex  figure, 
the  sect  AB,  joining  two  of  the  points  in  its  perimeter,  lying  without 
the  figure ;  then  if  the  re-entrant  portion  A  CB  be  revolved  about  the 
line  AB  into  the  position  AC'B,  the  figure  AC'BNA  has  the  same 
perimeter  as  the  first  figure,  but  a  greater  surface. 

Now  let  AFBCA  be  a  figure  of  greatest  surface  formed  with  a 
given  perimeter ;  then,  taking  any  point  A  in  its  perimeter,  and  drawing 


AB  to  bisect  the  perimeter,  it  also  bisects  the  surface.  For  if  the  sur- 
face of  one  of  the  parts,  as  AFB,  were  greater  than  that  of  the  other 
part,  A  CB,  then  if  the  part  AFB  were  revolved  upon  the  line  AB  into 
the  position  AF'B,  the  surface  of  the  figure  AF'BFA  would  be  greater 
than  that  of  the  figure  AFBCA,  and  yet  would  have  the  same  perim- 
eter. 

Now  the  angles  AFB  and  AF'B  must  be  right  angles,  else  the  tri- 
angles AFB  and  AF'B  could  be  increased,  by  443,  without  varying 
the  chords  AF,FB,  AF',F'B,  and  then  (the  segments  AGF,  FEB, 
AG'F' ,  F'E'B,  still  standing  on  these  chords)  the  whole  figure  would 
have  increased  without  changing  its  perimeter. 

But  F  is  any  point  in  the  curve  AFB  \  therefore  this  curve  is  a 
semicircle. 

(384.  The  arc  of  a  segment  is  a  semicircle  if  the  angle  in  it  is  right.) 

Therefore  the  whole  figure  is  a  circle. 


164 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  V. 

445*  Of  all  equivalent  plane  figures^  the  circle  has  the  least 
peri7neter. 


Hypothesis.    Lei  C  be  a  circle,  and  A  any  other  figure  having 
the  same  surface  as  C. 

Conclusion.    The  perimeter  of  C  is  less  than  that  of  A. 

Proof.     Suppose  B  a  circle  with  the  same  perimeter  as  the  figure 
A  ;  then,  by  444,  A  <B  :.     C  <B. 

But,  of  two  circles,  that  which  has  the  less  surface  has  the  less 
perimeter ; 

.-.     perimeter  of  C  <  perimeter  of  B,  or  of  A. 


Theorem  VI. 

446.  Of  all  the  polygons  constructed  with  the  same  given 
sides y  that  is  the  greatest  which  can  be  inscribed  in  a  circle. 


Hypothesis.     Lei  P  be   a  polygon   consirucied  wiih  ihe  sides 
a,  b,  c,  d,  e,  and  inscribed  in  a  circle  S,  and  lei  P'  be  any  oiher 


REGULAR  POLYGONS.  1 65 

polygon  constructed  with  the  same  sides,  and  not  inscriptible  in  a 
circle. 

Conclusion.    F  >  P'. 

Proof.  Upon  the  sides  a^  b,  c,  etc.,  of  the  polygon  P'  construct 
circular  segments  equal  to  those  standing  on  the  corresponding  sides  of 
P.  The  whole  figure  S'  thus  formed  has  the  same  perimeter  as  the 
circle  Sy  therefore,  by  444,  surface  oi  S  >  S' ;  subtracting  the  circular 
segments  from  both,  we  have 

P>P\ 


Theorem  VII. 

447.  Of  all  isoperimetric  polygons  having  the  sami  nttmber 
of  sides y  the  regular  polygon  is  the  greatest. 


Proof.  I.  The  greatest  polygon  P,  of  all  the  isoperimetric  poly- 
gons of  the  same  number  of  sides,  must  have  its  sides  equal ;  for  if  two 
of  its  sides,  as  AB' ,  B'C,  were  unequal,  we  could,  by  439,  increase  its 
surface  by  substituting  for  the  triangle  AB'C  the  isoperimetric  isosceles 
triangle  ABC. 

II.  The  greatest  polygon  constructed  with  the  same  number  of 
equal  sides  must,  by  446,  be  inscriptible  in  a  circle.  Therefore  it  is  a 
regular  polygon. 

Exercises,  ^y.  Of  all  triangles  that  can  be  inscribed  in  a 
given  triangle,  that  v^hose  vertices  are  the  feet  of  the  altitudes 
of  the  original  triangle  has  the  least  perimeter. 


i66 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   VIII. 

448.  Of  all  eqiiivalefit  polygons  having  the  same  number  of 
sides ^  the  regular polyg07i  has  the  least  perimeter. 


Hypothesis.  Let  P  be  a  regular  polygon,  and  M  any  equiva- 
lent irregular  polygon  having  the  same  number  of  sides  as  P. 

Conclusion.     The  perimeter  of  P  is  less  than  that  of  M. 
Proof.     Let  A"  be  a  regular  polygon  having  the  same   perimeter 
and  the  same  number  of  sides  as  M \  then,  by  447, 

M  <N,  or         P  <N. 

But,  of  two  regular  polygons  having  the  same  number  of  sides,  that 
which  has  the  less  surface  has  the  less  perimeter ;  therefore  the  perim- 
eter of  P  is  less  than  that  of  N  or  of  M. 


Theorem   IX. 

449.  If  a  regular  polygon  be  constructed  with  a  given  perim- 
eter, its  surface  will  be  the  greater^  the  greater  the  number  of  its 
sides. 


REGULAR  POLYGONS. 


167 


Proof.  Let  P  be  the  regular  polygon  of  three  sides,  and  Q  the 
regular  polygon  of  four  sides,  constructed  with  the  same  given  perim- 
eter. 

In  any  side  AB  of  P  take  any  arbitrary  point  D  \  the  polygon  P 
may  be  regarded  as  an  irregular  polygon  of  four  sides,  in  which  the 
sides  AD  J  DB,  make  a  straight  angle  with  each  other;  then,  by  447, 
the  irregular  polygon  P  of  four  sides  is  less  than  the  regular  isoperi- 
metric  polygon  Q  of  four  sides. 

In  the  same  manner  it  follows  that  Q  is  less  than  the  regular  isoper- 
imetric  polygon  of  five  sides,  and  so  on. 


Theorem  X. 

450.  Of  equivalent  regular  polygons^  the  perimeter  will  be 
the  less  J  the  greater  the  number  of  sides. 


p 

R 


Hypothesis.     Let  P  and  Q  be  equivalent  regular  polygons,  and 
let  Q  have  the  greater  number  of  sides. 

Conclusion.    The  perimeter  of  P  will  be  greater  than  that  of  Q. 
Proof.     Let  i?  be  a  regular  polygon  having  the  same  perimeter  as 
Q  and  the  same  number  of  sides  as  P;  then,  by  449, 

Q>P,        or        P>P; 

therefore  the  perimeter  of  P  is  greater  than  that  of  P  or  of  Q, 


BOOK   V. 


RATIO  AND   PROPORTION. 

Multiples. 

451.  Notation.     In  Book  V.,  capital  letters  denote  magni- 
tudes. 

Magnitudes  which   are   or  may  be   of   different   kinds   are 
denoted  by  letters  taken  from  different  alphabets. 
.^  +  ^  is  abbreviated  into  2A. 

A  +  A  +  A  =  sA. 

The  small  Italic  letters  w,  «,  /,  ^,  denote  whole  numbers. 

mA  means  A  taken  m  times ; 
nA  means  A  taken  n  times  j 
.*.     mA  +  nA  =  {m  +  n)A. 
nA  taken  m  times  is  mnA. 

452.  A  greater  magnitude  is  said  to  be  a  Multiple  of  a  lesser 

magnitude  when  the  greater  is  the  sum  of  a  number  of  parts 

each  equal  to  the  less ;  that  is,  when  the  greater  contains  the 

less  an  exact  number  of  times. 

Z69 


I/O  THE  ELEMENTS  OF  GEOMETRY. 

453.  A  lesser  magnitude  is  a  Submultiple^  or  Aliquot  Party 
of  a  greater  magnitude  when  the  less  is  contained  an  exact 
number  of  times  in  the  greater. 

454.  When  each  of  two  magnitudes  is  a  multiple  of,  or 
exactly  contains,  a  third  magnitude,  they  are  said  to  be  Com- 
mensurable. 

455.  If  there  is  no  magnitude  which  each  of  two  given  mag- 
nitudes will  contain  an  exact  number  of  times,  they  are  called 
Incommensurable. 

456.  Remark.  It  is  important  the  student  should  know, 
that  of  two  magnitudes  of  the  same  kind  taken  at  hazard,  or 
one  being  given,  and  the  other  deduced  by  a  geometrical  con- 
struction, it  is  very  much  more  likely  that  the  two  should  be 
incommensurable  than  that  they  should  be  commensurable. 

To  treat  continuous  magnitudes  as  commensurable  would 
be  to  omit  the  normal,  and  give  only  the  exceptional  case. 
This  makes  the  arithmetical  treatment  of  ratio  and  proportion 
radically  incomplete  and  inadequate  for  geometry. 


Problem   I. 

457.   To  find  the  greatest  common  submultiple   or  greatest 
common  divisor  of  two  given  magnitudeSy  if  any  exists. 


I 1 1 ^ 1 1 1  ^       J£JJ 


Let  AB  and  CD  be  the  two  magnitudes. 

From  AB,  the  greater,  cut  off  as  many  parts  as  possible,  each  equal 
to  CZ>y  the  less.  If  there  be  a  remainder  PB,  set  it  off  in  like  manner 
as  often  as  possible  upon  CZ>.  Should  there  be  a  second  remainder 
HD,  set  it  off  in  like  manner  upon  the  first  remainder,  and  so  on. 


RATIO  AND  proportion:  171 

The  process  will  terminate  only  if  a  remainder  is  obtained  which  is 
an  aliquot  part  of  the  preceding  one ;  and,  should  it  so  terminate,  the 
two  given  magnitudes  will  be  commensurable,  and  have  the  last  remainder 
for  their  greatest  common  divisor. 

For  suppose  HD  the  last  remainder. 

Then  HD  is  an  aliquot  part  of  FB,  and  so  of  CH,  and  therefore 
of  CD,  and  therefore  of  AF.  Thus  being  a  submultiple  of  AF  and 
FB,  it  is  contained  exactly  in  AB.  And,  moreover,  it  is  the  greatest 
common  divisor  of  AB  and  CD. 

For  since  every  divisor  of  CD  and  AB  must  divide  AF,  it  must 
divide  FB  or  CH,  and  therefore  also  HD. 

Hence  the  common  divisor  cannot  be  greater  than  HD. 

458.  Inverse  of  457.  If  two  magnitudes  be  commensura- 
ble, the  above  process  will  terminate. 

For  now,  by  hypothesis,  we  have  a  greatest  common  divisor 
G. 

But  G  is  contained  exactly  in  every  remainder. 

For  Gy  being  a  submultiple  of  CD,  is  also  an  aliquot  part 
of  AF,  a  multiple  of  CD ;  and  therefore,  to  be  a  submultiple  of 
AB,  it  must  be  an  aliquot  part  of  FB  the  first  remainder.  Sim- 
ilarly, CD  and  the  first  remainder  FB  being  divisible  by  G, 
the  second  remainder  HD  must  be  so,  and  in  the  same  way  the 
third  and  every  subsequent  remainder. 

But  the  alternate  remainders  decrease  by  more  than  half, 
and  so  the  process  must  terminate  at  G ;  for  otherwise  a 
remainder  would  be  reached  which,  being  less  than  G,  could 
not  be  divisible  by  G. 

459.  CoNTRANOMiNAL  OF  457.  The  above  process  applied 
to  incommensurable  magnitudes  is  interminable. 

460.  Obverse  of  457.  If  the  above  process  be  intermin- 
able, the  magnitudes  are  incommensurable. 

On  this  depends  the  demonstration,  given  in  461,  of  a 
remarkable  theorem  proved  in  the  tenth  book  of  Euclid's 
"  Elements." 


1/2 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   I. 
461.   The  side  and  diagonal  of  a  square  are  incommensurable. 


F  , 


Hypothesis.  Lei  ABCD  be  a  square;  AB,  a  side;  AC,  a 
diagonal. 

Conclusion.     Then  will  AB  and  AChQ  incommensurable. 
Proof.    AC>  AB,  but  <  2AB, 

Therefore  a  first  remainder  ^C  is  obtained  by  setting  off  on  AC  di 
part  AE  =  AB. 

Erect  EF  perpendicular  to  ^C  and  meeting  CD  in  F.    Join  AF. 

A  ADF  ^  A  AEF, 

(179.  Right  triangles  are  congruent  when  the  hypothenuse  and  one  side  are  equal 

respectively  in  each.) 

.\    DF  =  FE. 

Again,  rt.  A  CEF  is  isosceles,  because  one  of  the  complemental 
angles,  ECF,  is  half  a  rt.  ^, 

.;.     CE  =  EF  =  FD. 

Hence  a  common  divisor  oi  EC  and  DC  would  be  also  a  common 
divisor  oi  EC  and  EC. 

But  EC  and  EC  are  again  the  side  and  diagonal  of  a  square,  there- 
fore the  process  is  interminable. 


RATIO  AND  PROPORTION.  1 73 

462.  Another  demonstration  that  the  side  and  diagonal  of 
any  sqiiare  are  incommensurable. 

If  you  suppose  them  commensurable,  let  P  represent  their  common 
measure ;  that  is,  the  aliquot  part  which  is  contained  an  exact  number 
of  times  in  the  side,  and  also  an  exact  number  of  times  in  the  diagonal. 

Let  VI  represent  the  number  of  times  P  is  contained  in  the  side  6", 
and  n  the  number  of  times  P  is  contained  in  the  diagonal  D ;  so  that 
S  =  mP,  and  D  =  nP,  where  m  and  n  are  whole  numbers. 

Then,  by  243,  the  square  on  D  =  2  sq.  on  S,  and 

therefore  n^  is  an  even  number, 

therefore  n  is  an  even  number,  since  the  square  of  an  odd  number  is 

odd  j  therefore  n  may  be  represented  by  2^, 

.*.     2^^  =  m^, 

therefore  m  must  be  even,  and  can  be  represented  by  2t,  and  therefore 
^  must  be  even,  and  so  on  forever. 

That  is,  m  and  n  must  be  such  whole  numbers  that  they  will  divide 
by  2,  and  give  quotients  which  will  again  divide  by  z,  and  so  on  forever. 

This  is  impossible  ;  for  even  if  m  and  n  were  high  powers  of  2,  they 
would  by  constant  division  reduce  to  i,  an  odd  number. 


Scales   of   Multiples. 

463.  By  taking  magnitudes  each  equal  to  A,  one,  and  two, 
three,  four,  etc.,  of  them  together  we  obtain  a  set  of  magni- 
tudes depending  upon  A,  and  all  known  when  A   is  known;' 
namely.  A,  2^4,  $A,  4A,  sA,  6A,  .  .  .  and  so  on,  each  being 
obtained  by  putting  A  to  the  preceding  one. 

This  we  shall  call  the  scale  of  multiples  of  ^. 

464.  If  in  be  a  whole  number,  7nA  and  mBare  called  equi- 
multiples of  A  and  P,  or  the  same  multiples  of  ^and  B. 


1/4  THE  ELEMENTS   OF  GEOMETRY. 

465.  We  assume:  As  A  is  greater  than,  equal  to,  or  less 
than  By  so  is  mA  greater  than,  equal  to,  or  less  than  mB 
respectively. 

466.  By  33,  Rule  of  Inversion, 

As  mA  is  greater  than,  equal  to,  or  less  than  mB^  so  is  A 
greater  than,  equal  to,  or  less  than  B  respectively. 


Theorem  II. 

467.  Commensurable  magnitudes  have  also  a  common  mul- 
tiple. 

If  A  and  B  are  commensurable  magnitudes,  there  is  some  multiple 
of  A  which  is  also  a  multiple  of  B. 

Proof.     Let  C  be  a  common  divisor  of  A  and  B. 
The  scale  of  multiples  of  C  is 

C,  2C,  3C  .... 

Now,  by  hypothesis,  one  of  the  multiples  of  this  scale,  suppose  pC, 
is  equal  to  A,  and  one,  suppose  qC,  is  equal  to  B. 

Hence,  by  465,  the  multiple  pqC  is  equal  to  qAy  and  the  same 
multiple  is  equal  to  pB  \  therefore, 

qA  =  pB. 

468.  Inverse  of  467.  Magnitudes  which  have  a  common 
multiple  are  commensurable. 

Proof.  If  pA  =  qB,  then  —  will  go  p  times  into  B,  and  q  times 
into  A. 

469.  Any  whole  number  or  fraction  is  commensurable  with 
every  whole  number  and  fraction,  being  each  divisible  by  unity 
over  the  product  of  the  denominators. 

To  find  a  common  multiple,  we  have  only  to  multiply 
together  the  whole  numbers  and  the  numerators  of  the  frac- 
tions. 


RATIO  AND  PROPORTION.  1 75 

470.  Multiplication  by  a  whole  number  or  fraction  is  dis- 
tributive, 

m{A  ■\-  B  -\-  .  .  ,  )  =^  mA  -^  vtB  -^  ,  .  . 

471.  Multiplication  being  commutative  for  whole  numbers 
or  fractions, 

.*.     m{nA)  =  mnA  —  nmA  =  n{mA). 

472.  Magnitudes  which   are  of   the   same  kind  can,  being 
multiplied,  exceed  each  the  other. 


Scale   of   Relation. 

473.  The  Scale  of  Relation  of  two  magnitudes  of  the  same 
kind  is  a  list  of  the  multiples  of  both,  all  arranged  in  ascending 
order  of  magnitude ;  so  that,  any  multiple  of  either  magnitude 
being  assigned,  the  scale  of  relation  points  out  between  which 
multiples  of  the  other  it  lies. 

474.  If  we  call  the  side  of  any  square  5,  and  its  diagonal 
D,  their  scale  of  relation  will  commence  thus,  — 

S,  D,  2S,  2D,  zS,  46",  3A  5^,4A  65,  76*,  5 A  ZS,  6D,  gS,  jD, 
loS,  iiS,  SZ>,  12S,  gD,  135,  14S,  loDj  156",  iiX>,  166*,  12I?,  176", 
186",  i^D,  igS,  i^D,  20S,  21S,  !$£>,  etc. 

141^,  looD,  142S,  etc. 

14142136",  iooooooZ>,  14142146,  etc. 

1414213566',  iooooooooZ>,  1414213576,  etc. 

And  we  have  proved  that  no  multiple  of  5  will  ever  equal 
any  multiple  of  D. 

475.  If  A  be  less  than  B,  one  multiple  at  least  of  the  scale 
of  A  will  lie  between  each  two  consecutive  multiples  of  the 
scale  of  B. 

Moreover,  if  A  and  B  are  two  finite  magnitudes  of  the  same 


1/6  THE  ELEMENTS  OF  GEOMETRY. 

kind,  however  small  A  may  be,  we  may,  by  continuing  the  scale 
of  multiples  of  A  sufficiently  far,  at  length  obtain  a  multiple 
of  A  greater  than  B. 

So  we  are  justified  in  saying, 
I.  We  can  always  take  mA  greater  than  B  or  pB. 

II.  We  can  always  take  nA  such  that  it  is  greater  than/^, 
but  not  greater  than  qB,  provided  that  A  is  less  than  B,  and  / 
than  q. 

476.  The  scale  of  relation  of  two  magnitudes  will  be  changed 
if  one  is  altered  in  size  ever  so  little ;  for  some  multiple  of  the 
altered  magnitude  can  be  found  which  will  exceed,  or  fall  short 
of,  the  same  multiple  of  it  before  alteration,  by  more  than  the 
other  original  magnitude,  and  consequently  the  interdistribu- 
tion  of  the  multiples  of  the  two  original  magnitudes  will  differ 
from  the  interdistribution  of  the  multiples  of  one  of  the 
original  magnitudes  and  the  second  altered. 

Hence,  when  two  magnitudes  are  known,  the  order  of  their 
multiples  is  fixed  and  known. 

Inversely,  if  by  any  means  the  order  of  the  multiples  is 
known,  and  also  one  of  the  original  magnitudes,  the  other  is  of 
fixed  size,  even  though  we  may  not  yet  be  in  a  condition  to 
find  it. 

477'  The  order  in  which  the  multiples  of  A  lie  among  the 
multiples  of  B,  when  all  are  arranged  in  ascending  order  of 
magnitude,  and  the  series  of  multiples  continued  indefinitely, 
determines  what  is  called  the  Ratio  of  A  to  B,  and  written 
A  :B,in  which  the  first  magnitude  is  called  the  Antecedent,  and 
the  second  the  Consequent. 

478.  Hitherto  we  have  compared  one  magnitude  to  another, 
with  respect  to  quantity,  only  in  general,  according  to  the 
logical  division  greater  than,  equal  to,  less  than. 

But  double,  triple,  quadruple,  quintuple,  sextuple,  are  spe- 
cial cases  of  a  more  subtile  relation  which  exists  between 
every  two  magnitudes  of  the  same  kind. 


RATIO  AND   proportion:  1 7/ 

Ratio  is  the  relation  of  one  magnitude  to  another  with 
respect  to  quantuplicity. 

479.  If  the  multiples  of  A  and  B  interlie  in  the  same  order 
as  the  multiples  of  C  and  D,  then  the  ratio  ^  to  ^  is  the  same 
as  the  ratio  C  to  Dy  and  the  four  magnitudes  are  said  to  be 
Proportional^  or  to  form  a  Proportion. 

This  is  written  A  :  B  \:  C  :  D,  and  read  "  ^  to  ^  is  the 
same  as  ^'to  />." 

The  second  pair  of  magnitudes  may  be  of  a  different  kind 
from  the  first  pair. 

A  and  D  are  called  the  Extrejues,  B  and  C  the  Means ^  and 
D  is  said  to  be  a  Fourth  Proportional  to  A^  By  and  C. 

480.  In  the  special  case  when  A  and  B  are  commensurable, 
we  can  estimate  their  quantuple  relation  by  considering  what 
multiples  they  are  of  some  common  standard  ;  and  so  we  can  get 
two  numbers  whose  ratio  will  be  the  same  as  the  ratio  of  A  to  B. 

481.  The  ratio  of  two  magnitudes  is  the  same  as  the  ratio 
of  two  other  magnitudes,  when,  any  equimultiples  whatsoever 
of  the  antecedents  being  taken,  and  likewise  any  equimultiples 
whatsoever  of  the  consequents,  the  multiple  of  one  antecedent 
is  greater  than,  equal  to,  or  less  than,  that  of  its  consequent, 
according  as  the  multiple  of  the  other  antecedent  is  greater 
than,  equal  to,  or  less  than,  that  of  its  consequent. 

482.  Thus,  the  ratio  oi  A  \.o  B  is  the  same  as  the  ratio  of  C 
to  D  when  mA  is  greater  than,  equal  to,  or  less  than  nBy 
according  as  mC  is  greater  than,  equal  to,  or  less  than  nBy 
whatever  whole  numbers  m  and  n  may  be. 

483.  Three  magnitudes  (Ay  By  C)  of  the  same  kind  are  said 
to  be  proportionals  when  the  ratio  of  the  first  to  the  second  is 
the  same  as  the  ratio  of  the  second  to  the  third ;  that  is,  when 

A  \  B  '.'.  B  \  C. 

In  this  case,  C  is  said  to  be  the  Third  Proportional  to  A  and 
By  and  B  the  Mean  Proportional  between  A  and  C. 


178  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   III. 

484.  Ratios  which  are  the  same  as  the  same  ratio  are  the 
same  as  07te  ajiother. 

Hypothesis.    A  \  B  w   C  \  D^'axA  A  \  B  w  X  -.   V. 

Conclusion.     C  \  D  w  X  \   V. 

Proof.  By  the  inverse  of  479,  the  multiples  of  C  and  D  have  the 
same  interorder  as  those  of  A  and  B ;  and,  the  same  being  true  of 
the  multiples  of  X  and  V,  therefore  the  multiples  of  X  and  V  have  the 
same  interorder  as  those  of  C  and  Z). 

485.  Remark.  Thus  ratios  come  under  the  statement, 
"Things  equal  to  the  same  thing  are  equal  to  each  other;" 
and  a  proportion  may  be  spoken  of  as  an  equality  of  ratios. 

486.  Since  the  inverse  of  a  complete  definition  is  true, 
therefore  if  a  pair  of  numbers  /  and  ^  can  be  found  such  that 
pA  is  greater  than,  equal  to,  or  less  than  gB  when  pC  is  not 
respectively  greater  than,  equal  to,  or  less  than  gDy  then  the 
ratio  oi  A  to  B  is  unequal  to  the  ratio  of  C  to  D. 

487.  The  first  is  to  be  to  the  second  in  a  greater  (or  less) 
ratio  than  the  third  to  the  fourth  when  a  multiple  of  the  first 
takes  a  more  (or  less)  advanced  position  among  the  multiples 
of  the  second  than  the  same  multiple  of  the  third  takes  among 
those  of  the  fourth. 

488.  The  ratio  of  ^  to  ^  is  greater  than  that  oi  C  to  D 
when  two  whole  numbers  m  and  n  can  be  found  such  that  mA 
is  greater  than  nB,  while  7nC  is  not  greater  than  nD ;  or  such 
that  mA  is  equal  to  nB,  while  mCis  less  than  nD. 


Theorem  IV. 

489.  Equal  magnitudes  have  the  same  ratio  to  the  same 
magnitude^  and  the  same  has  the  same  ratio  to  equal  magni- 
tudes. 


RATIO  AND  proportion:  1 79 


For  if  A  =  B, 

.*.     mA  =  mB. 

So  if  mA  >  nC, 

.*.     mB  >  nC; 

and  if  equal,  equal ;  if  less,  less ; 

/.     A  :   C  ::  B  :   a 
And  also  if  ^T  >  mA, 

.'.     pT>  mB; 
and  if  equal,  equal ;  if  less,  less ; 

/.     T  :  A  ::  T  ',  B, 


Theorem  V. 

490.  Of  two  uneqtial  magnitudes,  the  greater  has  a  greater 
ratio  to  any  other  magnitude  than  the  less  has ;  and  the  same 
magnitude  has  a  greater  ratio  to  the  less,  of  two  other  magni- 
tudes, tlian  it  has  to  the  greater, 

\i  A  >  B,m  can  be  found  such  that  mB  is  less  than  mA  by  a 
greater  magnitude  than  C. 

Hence,  if  mA  —  nC,  or  if  mA  is  between  wCand  {n  -f-  1)6',  mB 
will  be  less  than  nC \  therefore,  by  488, 

A  '.   C  >  B  :   C, 
Also,  since  nC  >  mB,  while  « C  is  not  >  mA, 
/.     C  :  B  >  C  :  A, 

491.  From  489  and  490,  by  33,  Rule  of  Inversion, 

If  ^  :  C  ::  ^  :  C,  or  if  T  :  A  ::  T  :  B, 

:.  A  =  B. 
li  A  :C>B:C,  or  a  T  :A<T  :B, 

.-.  A  >  B. 
If  A  :  C<B  :C,oriiT:A>T:B, 

/.    A  <  B. 


l80  THE  ELEMENTS  OF  GEOMETRY, 


Theorem  VI. 

492.  If  any  number  of  magnitudes  he  proportionals y  as  one 
of  the  antecedents  is  to  its  consequent^  so  will  all  the  antecedents 
taken  together  be  to  all  the  consequents. 

Let 

A  ',  B  ',',  C  ',  D  w  E  '.  F, 
then 

A  \  B  '.'.  A  -ir  C  ^  E  ',  B  -\-  D  -ir  F, 

For  as  mA  >,  =,  or  <  nB^  so,  by  inverse  of  482,  is  »?C  >,  =,  or 
<  nD  \  and  so  also  is  mE  >,  =,  or  <  nF; 

.',     so  also  is  mA  +  mC  -{•  mE  >,  =,  or  <  nB  +  nD  +  nF, 

and  therefore  so  is  m{A  -\-  C  +  E)  >,  —,  or  <  n{B  +  D  ■}-  F); 

whence 

A  :  B  ::  A  +  C  +  E  :  B  -\-  Z>  -i-  F. 


Theorem  VII. 

493-   The  ratio  of  the  equimultiples  of  two  magnitudes  is  the 
same  as  the  ratio  of  the  magnitudes  themselves, 

mA  \  mB  w  A  \  B, 

For  as  pA  >,  =,  or  <  qB,  so  is  mpA  >,  =,  or  <  mqB ', 

but  mpA  =  pmA,  and  mqB  =  qmBy 

.*.     as  pA  >,  =,  or  <  qB, 

so  is  pmA  >,  =,  or  <  qmB,  whatever  be  the  values  of  p  and  q; 

.-.    A  :  B  ::  mA  :  mB. 


RATIO  AND  proportion:  i8i 


Theorem  VIII. 

494.  If  four  magnitudes  be  proportionals  y  then^  if  the  first  be 
greater  than  the  third,  the  second  will  be  greater  than  the  fourth; 
and  if  equal,  equal ;  and  if  less,  less. 

Given,  A  \  B   \\  C  \  D. 

\i  A  >  C, 

/.     A  '.  B  >   C  :  B; 

therefore,  from  our  hypothesis, 

C  I  r>  >   C  '.  B; 


therefore,  by  491, 
If  ^  =  C, 


If  ^  <  C, 


B  >  D. 

,'.     A  :  B  :i  C  :  B, 
,\     C  :  V  ::  C  :  B, 

.-.   B  =^n, 

/.     A  :  B  <   C  :  By 

/.     C  '.  D  <   C  ;  B, 

/.    B  <  D, 

Theorem  IX. 

495'  If  four  magnitudes  of  the  same  kind  be  proportionals, 
they  will  also  be  proportionals  when  taken  alternately. 

Let  A  \  B  w   C  :  /?,  the  four  magnitudes  being  of  the  same  kind, 

then  alternately, 

A  '.   C  :.  B  \  D. 
For,  by  493, 

mA\  mB  \\  A  \  B  '.:   C  .  D  \\  nC  ',  nD, 

.*.     mA  :  mB  \\  nC  \  nD \ 
therefore,  by  494,  mA  >,  =,  or  <  «C,  as  mB  >,  =,  or  <  nD',  and,  this 
being  true  for  all  values  of  m  and  «, 

.-.       A  \  C  ','.  B  \  D, 


1 82  THE  ELEMENTS  OF  GEOMETRY, 


Theorem   X. 

496.  If  two  ratios  are  equals  the  sum  of  the  antecedent  and 
consequent  of  the  first  has  to  the  consequent  the  same  ratio  as 
the  sum  of  the  antecedent  and  consequent  of  the  other  has  to  its 
consequent. 

li  A  :  B  ::  C  '.  D, 
then 

A  ^  B  '.  B  ','.  C  +  D  :  D, 

Proof.  Whatever  multiple  of  ^  +  ^  we  choose  to  examine,  take 
the  same  multiple  of  A,  say  1 7^,  and  let  it  lie  between  some  two  mul- 
tiples of  Bf  say  23^  and  24^ ;  then,  by  hypothesis,  1 7  C  lies  between 
23Z)  and  24D. 

Add  I  "jB  to  all  the  first,  and  i  "jD  to  all  the  second ;  then  1 7  (^  + 
B)  lies  between  40^  and  41^,  and  i7(C  +  D)  between  40Z)  and 
412? ;  and  in  the  same  manner  for  any  other  multiples. 


BOOK   VI. 


RATIO  APPLIED. 
I.   Fundamental  Geometric  Proportions. 

Theorem   I. 

497*  J^f  i'^o  ^i^^s  ^^^  ^^^  by  three  parallel  lines ^  the  intercepts 
on  the  one  are  to  07ie  aiiother  in  the  same  ratio  as  the  correspond- 
ing intercepts  on  the  other. 


Hypothesis.     Lei  the  three  parallels  AA',  BB\  CC,  cut  two 
other  lines  in  A^  B,  C,  and  A',  B\  C ,  respectively. 


183 


1 84 


THE  ELEMENTS  OF  GEOMETRY. 


Conclusion.     AB  :  BC  ::  A'B'  :  B' C , 

Proof.  On  the  line  ABC,  by  laying  off  m  sects  =  AB,  take  BM 
=  mAB,  and,  in  the  same  way,  BN  =  n.BC,  taking  J/ and  N  on  the 
same  side  of  B.  From  J/  and  N  draw  lines  1|  AA',  cutting  A'B'C 
in  J/'  and  iV^'. 

.-.    B'M'  =^  m,A'B\         and        B'N'^n.B'C. 

(227.  If  three  or  more  parallels  intercept  equal  sects   on  one   transversal,  they 

intercept  equal  sects  on  every  transversal. ) 

But  whatever  be  the  numbers  m  and  n,  as  BM  (or  m .  AB)  is  >, 
=  ,  or  <BN  {orn.BC), 
so  is  B'M'  (or  w  .  A'B')  respectively  >,  =,  or  <  B'N"  (or  n .  ^'C')  ; 

.-.     AB  :  BC  ::  A'B'  :  ^'6". 

498.  Remark.  Observe  that  the  reasoning  holds  good, 
whether  B  is  between  A  and  C,  or  beyond  A^  or  beyond  C, 


499.  Corollary  I.  If  the  points  A  and  ^4'  coincide,  the 
figure  ACC  will  be  a  triangle ;  therefore  a  line  parallel  to  one 
side  of  a  triangle  divides  the  other  two  sides  proportionally. 

500.  Corollary  II.  If  two  lines  are  cut  by  four  parallel 
lines,  the  intercepts  on  the  one  are  to  one  another  in  the  same 
ratio  as  the  corresponding  intercepts  on  the  other. 

501.  If  a  sect  AB  is  produced,  and  the  line  cut  at  a  point  P 
outside  the  sect  AB,  the  sect  AB  is  said  to  be  divided  exter- 
nally at  jP,  and  AP  and  BP  are  called  External  Segments  of 
AB, 


RATIO  APPLIED.  1 85 


In  distinction,  if  the  point  P  is  on  the  sect  AB^  it  is  said 
to  be  divided  ifiternally. 


Theorem  II. 

502.  A  given  sect  can  be  divided  internally  into  two  seg- 
ments having  the  same  ratio  as  any  two  given  sectSy  and  also 
externally  unless  the  ratio  be  07te  of  equality  ;  aftdj  in  each  case, 
there  is  only  o?te  such  point  of  division. 


I   jr  G 

:b 

-N   M 

\ 

^^4 

\\ 

w 

\ 

% 

^.  1 

\ 
\ 
\ 

\ 

\ 

\l 

\ 

JC^ 

Given,  ihe  sect  AB. 

On  a  line  from  A  making  any  angle  with  AB,  take  A  C  and  CD 
equal  to  the  two  given  sects.     Join  BD. 

Draw  CF  \\  DB  and  meeting  AB  in  F.     By  497,  AB  is  divided 
internally  at  F  in  the  given  ratio. 

If  it  could  be  divided  internally  at  G  in  the  same  ratio,  BH  being 
drawn  ||    CG  to  meet  AD  in  H, 
A  G  would  be  to  GB  as  ^  C  to  CH,  and  therefore  not  as  ^  C  to  CD. 

(476.  The  scale  of  relation  of  two  magnitudes  will  be  changed  if  one  is  altered  in 

size  ever  so  little.) 

Hence  F  is  the  only  point  which  divides  AB  internally  in  the  given 
ratio. 


i86 


THE  ELEMENTS  OF  GEOMETRY, 


Again,  if  CD  be  taken  so  that  A  and  D  are  on  the  same  side  of  C, 
the  like  construction  will  determine  the  external  point  of  division. 

In  this  case  the  construction  will  fail  if  CD  =  AC,  for  D  would 
coincide  with  A. 

As  above,  we  may  also  prove  that  there  can  be  only  one  point  of 
external  division  in  the  given  ratio. 


C\ 


503.  Inverse  of  499.  A  line  which  divides  two  sides  of  a 
triangle  proportionally  is  parallel  to  the  third. 

For  a  parallel  from  one  of  the  points  would  divide  the 
second  side  in  the  same  ratio,  but  there  is  only  one  point  of 
division  of  a  given  sect  in  a  given  ratio. 

Theorem   III. 

504.  Rectangles  of  equal  altitude  are^  to  one  another  in  the 
same  ratio  as  their  bases. 


: 1 


B    M 


j)r 


Let  AC,  BC,  be  two  rectangles  having  the  common  side  OC,  and 
their  bases  OA,  OB,  on  the  same  side  of  OC. 

In  the  line  OAB  take  OM  =  m  .OA,  and  ON  =  n.OB,  and 
complete  the  rectangles  MC  and  NC. 


RATIO  APPLIED.  1 87 


Then  MC  =  m.AC,  and  JVC  =  n.BC;  and  as  OM  is  >,  =,  or 
<  ON,  so  is  MC  respectively  >,  =,  or  <  NC ; 

.*.     rectangle  ^C  :  rectangle  ^C  ::  base  OA  :  base  OB. 

505.  Corollary.     Parallelograms  or  triangles  of  equal  alti- 
tude are  to  one  another  as  their  bases. 


Theorem  IV. 

506.  In  the  same  circle,  or  in  equal  circles,  angles  at  the  cen- 
ter and  sectors  are  to  one  another  as  the  arcs  on  which  they  stand. 


Let  O  and  C  be  the  centers  of  two  equal  circles ;  AB,  KL,  any 
two  arcs  in  them. 

Take  an  arc  AM  =  m  .  AB ; 
then  the  angle  or  the  sector  between  OA  and  OM  equals  m .  A  OB. 

(365.  In  equal  circles,  equal  arcs  subtend  equal  angles  at  the  center.) 

Also  take  an  arc  KN  =  n  .  KL ;  then  the  angle  or  sector  between 
CK  and  CN  equals  n  .  KCL. 

But  as  AOM>,  =,  or  <  KCN,  so  respectively  is  arc  AM  >,  =,  or 
<  arc  KN) 

(370  and  372.  In  equal  circles,  equal  angles  at  the  center  or  equal  sectors  intercept 
equal  arcs,  and  of  two  unequal  angles  or  sectors  the  greater  has  the  greater  arc.) 

.-.    AOB  :  KCL  ::  aicAB  :  aic  KL. 


1 88  THE  ELEMENTS  OF  GEOMETRY, 


II.   Similar  Figures. 

507.  Similar  figures  are  those  of  which  the  angles  taken  in 
the  same  order  are  equals  and  the  sides  between  the  equal  angles 
proportional 


The  figure  ABCD  is  similar  to  A'B'C'jy  \i  4.  A  ^  4.  A\  4.  B  ^ 
4.  B\  etc.,  and  also 

AB  :  A'B'  ::  BC  :  B' C  x\  CD  \   CU  : :  etc. 

Theorem  V. 
508.  Mutually  equiangular  triangles  are  similar. 


Hypothesis,     t^s  ABC,  FGH,  having  ^.s  at  A,  B,  and  C,  =  ^.s 
at  F,  G,  and  H. 

Conclusion.    AB  \  FG  w  BC  \  GH  ..  CA  -.  HF, 
Proof.    Apply  A  FGH  to  A  ABC  so  that  the  point  G  coincides 
with  B,  and  GH  falls  on  BC 
Then,  because  ^  6^  =  ^  ^, 

.-.     GF  faUs  on  BA. 


RATIO  APPLIED.  1 89 


Now,  since  2<  BF'H'  =  ^  A  by  hypothesis, 
.-.     J^'H'  II  AC, 
(166.  If  corresponding  angles  are  equal,  the  lines  are  parallel.) 

therefore,  by  498,  AB  :  BF'    -.:   CB  :  BH\ 

In  the  same  way,  by  applying  the  A  FGH  so  that  the  ^  s  at  ZT  and 
C  coincide,  we  may  prove  that  BC  \   GH  : :   CA  :  HF, 

509.  Corollary.     A  triangle  is  similar  to  any  triangle  cut 
off  by  a  line  parallel  to  one  of  its  sides. 


Theorem  VI. 

510.   Triangles  having  their  sides  taken  in  order  proportional 
are  similar. 


T  J£ 


Hypothesis.    AB  :  FG  ::  BC  -.  GH  -,:   CA  :  HF, 
Conclusion.     :^  C  =  ^  H,  and  -4.  A  —  -i^  F. 
Proof.     On  BA  take  BF"  =  GF,  and  draw  FH'  \\   CA ; 
.-.     AB  :  F'B  '.'.  BC  '.  BH'  : :   CA  :  HF'. 
(508.  Mutually  equiangular  triangles  are  similar.) 

Since  FG  =  F'B,        :,    BC  -.  GH  ::  BC  -.  BH\ 
(484.  Ratios  equal  to  the  same  ratio  are  equal.) 

therefore,  by  491,  GH  =  BH'. 

In  the  same  way  HF  =  H'F', 

.-.     L.FGH^  t.F'BH'. 
But  F'BH'  is  similar  to  ABC. 


1 90  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  VII. 

511.  Two  triangles  having  one  angle  of  the  one  equal  to  one 
angle  of  the  othevy  and  the  sides  about  these  angles  proportional, 
are  similar. 


Hypothesis.     ^  B  =  ^  G,  and  AB  :  BC  ..  FG  :  GH, 

Conclusion,     a  AB  C  ~  a  FGH  (using  ^  for  the  word  "  similar  ") . 
Proof.     In  BA  take  BF'  =  GF,  and  draw  F'L  \\  AC, 

.'.     A  F'BL  -  A  ABC, 
(508.  Equiangular  triangles  are  similar.) 

.-.    AB  :  BC  ::  F'B  :  BL. 

But  F'B  —  FG  by  construction ;  therefore,  from  our  hypothesis, 

AB  ',  BC  '.'.  F'B  :   GH, 

/.   FB  X  BL   i:  F'B  :  GH, 
therefore,  by  491, 

BL  =  GH, 

.-.     A  F'BL  ^  A  FGH, 

(124.  Triangles  having  two  sides  and  the  included  angle  respectively  equal  are 

congruent.) 

.-.     A  FGH  -  A  ABC. 


RATIO  APPLIED.  IQI 


Theorem  VIII. 

512.  If  two  triangles  have  two  sides  of  the  07ie  proportional 
to  two  sides  of  the  other^  and  an  angle  in  each  opposite  one  cor- 
responding pair  of  these  sides  equals  the  angles  opposite  the  other 
pair  are  either  equal  or  supplemental. 


The  angles  included  by  the  proportional  sides  are  either 
equal  or  unequal. 

Case  I.     If  they  are  equal,  then  the  third  angles  are  equal. 

(174.  The  sum  of  the  angles  of  a  triangle  is  a  straight  angle.) 

Case  II.     If  the  angles  included  by  the  proportional  sides 
are  unequal,  one  must  be  the  greater. 

Hypothesis.    ABC  and FGH  as  with  AB  \  BC  w  FG  \  GH, 
^A==^F,  ^B>^G, 

Conclusion.     ^  C  -f-  ^  ^  =  st.  ^ . 
Proof.     Make  ^  ABV  =  ^  6^ ; 

.♦.     A  ABD  -  A  FGH; 
(508.  Equiangular  triangles  are  similar.) 

.-.    AB  :  BD  ::  FG  :   GH; 
therefore,  from  our  hypothesis, 

AB  :  BV  ::  AB  :  BC] 

therefore,  by  491,  BD  =  BC, 

.*.     ^  C  =  ^  BJDC. 
But  ^  BDC  +  ^  BDA  =  st.  ^, 

.-.     ^  C  +  ^  Zr  =  St.  ^. 


192  THE  ELEMENTS  OF  GEOMETRY. 

513.  Corollary.  If  two  triangles  have  two  sides  of  the 
one  proportional  to  two  sides  of  the  other,  and  an  angle  in  each 
opposite  one  corresponding  pair  of  these  sides  equal,  then  if 
one  of  the  angles  opposite  the  other  pair  is  right,  or  if  they  are 
oblique,  but  not  supplemental,  or  if  the  side  opposite  the  given 
angle  in  each  triangle  is  not  less  than  the  other  proportional 
side,  the  triangles  are  similar. 

514.  In  similar  figures,  sides  between  equal  angles  are  called 
Homologous^  or  corresponding.  The  ratio  of  a  side  of  one 
polygon  to  its  homologous  side  in  a  similar  polygon  is  called 
the  Ratio  of  Similitude  of  the  polygons.  Similar  figures  are 
said  to  be  similarly  placed  vAv^n  each  side  of  the  one  is  parallel 
to  the  corresponding  side  of  the  other. 


Theorem   IX. 

515.  If  two  unequal  similar  figures  are  similarly  placed^  all 
lines  joining  a  vertex  of  one  to  the  corresponding  vertex  of  the 
other  are  concurrent. 


Hypothesis.    ABCD  and  A'B'C'iy  two  similar  figures   simi- 
larly placed. 

Conclusion.    The  lines  AA\  BB\  CC,  etc.,  meet  in  a  point  P. 
Proof.     Since  AB  and  A'B'  are  unequal, 

.-.    AA'  and  BB'  are  not  |1  . 
Call  their  point  of  intersection  F,    Then 

AF  :  A'F  '.:  BF  '.  B'F  ::  AB  :  A'B, 
(508.  Equiangular  triangles  are  similar.) 


RATIO  APPLIED.  1 93 


In  the  same  way,  if  BB'  and  CC  meet  in  Q,  then 

BQ  :  B'Q    ::  BC  :  B' C\ 

But,  by  hypothesis, 

AB   :  A'B'  '.',  BC  '.  B' C\ 

.-.     BP  :  B'P    ::  BQ  :  B' Q, 

.'.    the  point  Q  coincides  with  P. 

(502.  A  line  can  be  cut  only  at  a  single  point  into  external  segments  having  a  given 

ratio.) 

516.  Corollary.  Similar  polygons  may  be  divided  into 
the  same  number  of  triangles  similar  and  similarly  placed. 

For  if,  with  their  corresponding  sides  parallel,  one  of  the 
polygons  were  placed  inside  the  other,  the  lines  joining  cor- 
responding vertices  would  so  divide  them. 

517.  The  point  of  concurrence  of  the  lines  joining  the  equal 
angles  of  two  similar  and  similarly  placed  figures  is  called  the 
Center  of  Similitude  of  the  two  figures. 


518.  Corollary.  The  sects  from  the  center  of  similitude 
along  any  line  to  the  points  where  it  meets  corresponding  sides 
of  the  similar  figures  are  in  the  ratio  of  those  sides. 

Exercises,  %'j.  Construct  a  polygon  similar  to  a  given 
polygon,  the  ratio  of  similitude  of  the  two  polygons  being 
given. 


194 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   X. 

519.  y4  perpendicular  from  the  right  angle  to  the  hypothenuse 
divides  a  right-angled  triangle  into  two  other  triangles  similar 
to  the  whole  and  to  one  another. 


Hypothesis,     a  ABC  right-angled  at  C. 
CD  ±  AB. 

Conclusion,     t.  ACD  ^  t.  ABC  -^  t.  CBD. 
Proof.     ^  CAD  =  ^BAC,  and  rt.  ^  ADC  =  rt.  ^  ACB, 
.-.     ^ACD  =  ^ABC, 
(174.  The  three  angles  of  any  triangle  are  equal  to  a  straight  angle.) 

.-.     aACD  ~  hABC. 
In  the  same  way  we  may  prove  A  CBD  ~  A  ABC; 
.-.     aACD  ~  A  CBD. 

520.  Corollary  I.  Each  side  of  the  right  triangle  is  a 
mean  proportional  between  the  hypothenuse  and  its  adjacent 
segment. 

For  since  A  A  CD  ~  A  ABC, 

.-.     AB  :  AC  ::  AC  :  AD. 


521.  Corollary  II.  The  perpendicular  is  a  mean  propor- 
tional between  the  segments  of  the  hypothenuse. 

522.  Corollary  III.  Since,  by  383,  lines  from  any  point 
in  a  circle  to  the  ends  of  a  diameter  form  a  right  angle,  there- 
fore, if  from  any  point  of  a  circle  a  perpendicular  be  dropped 


RATIO  APPLIED. 


195 


upon  a  diameter,  it  will  be  a  mean  proportional  between  the 
segments  of  the  diameter. 


Theorem   XL 

523.  The  bisector  of  an  interior  or  exterior  angle  of  a  triangle 
divides  the  opposite  side  internally  or  externally  in  the  ratio  of 
the  other  two  sides  of  the  triangle. 


Hypothesis.    ABC  any  A.    BD  ihe  bisector  of  4.  at  B. 

Conclusion.    AB  :  BC  .'.  AD  \  DC. 

Proof.     Draw  AF  \\  BD. 

Then,  of  the  two  angles  at  B  given  equal  by  hypothesis,  one  equals 
the  corresponding  interior  angle  at  F,  and  the  other  the  corresponding 
alternate  angle  at  A, 

(168.  A  line  cutting  two  parallels  makes  alternate  angles  equal,  and  (169)  corre- 
sponding angles  equal.) 

.-.     AB  =  BF. 
(126.  If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  are  equal.) 

But  BF  :  BC  :-.  AD  -.  DC, 

(499.  If  a  line  be  parallel  to  a  side  of  a  triangle,  it  cuts  the  other  sides  propor- 
tionally.) 

/.    AB  :BC::ADi  DC, 


196  THE  ELEMENTS  OF  GEOMETRY. 

524.  Inverse.  Since,  by  502,  a  sect  can  be  cut  at  only  one 
point  internally  in  a  given  ratio,  and  at  only  one  point  exter- 
nally in  a  given  ratio,  therefore,  by  32,  Rule  of  Identity,  if  one 
side  of  a  triangle  is  divided  internally  or  externally  in  the  ratio 
of  the  other  sides,  the  line  drawn  from  the  point  of  division  to 
the  opposite  vertex  bisects  the  interior  or  exterior  angle. 

525.  When  a  sect  is  divided  internally  and  externally  into 
segments  having  the  same  ratio,  it  is  said  to  be  divided  har- 
monically. 

526.  Corollary.  The  bisectors  of  an  interior  and  exterior 
angle  at  one  vertex  of  a  triangle  divide  the  opposite  side  har- 
monically. 


Theorem   XII. 

527.  If  a  secty  AB,  is  divided  harmonically  at  the  points  P 
and  Q,  the  sect  PQ  will  be  divided  harmonically  at  the  points 
A  and  B. 

I ! , , 

3L  ^       7»  a 


Hypothesis.  THq  sect  AB  divided  internally  at  F,  and  exter- 
nally at  Q,  so  that 

AP  '.  BP  '.'.  AQ  :  BQ', 

therefore,  by  inversion, 

BP  '.  AP  '.'.  BQ  '.  AQ; 
therefore,  by  alternation, 

BP  :  BQ  '.:  AP  :  AQ. 

528.  The  points  A,  B,  and  P,  Q,  of  which  each  pair  divide 
harmonically  the  sect  terminated  by  the  other  pair,  are  called 
four  Harmonic  Points. 


RATIO  APPLIED, 


197 


III.   Rectangles  and  Polygons. 

Theorem   XIII. 

529.  If  four  sects  are  proportionaly  the  rectangle  contained  by 
the  extremes  is  equivalent  to  the  rectangle  contained  by  the 
means. 


ac 


ad 


^ 


Let  the  four  sects  a^  b,  c,  ^/,  be  proportional. 

Then  rectangle  ad  —  be. 

Proof.     On  a  and  on  b  construct  rectangles  with  altitude  =  c. 

On  c  and  on  d  construct  rectangles  of  altitude  a. 

Then 

a  :  b  ::  ac  :  bCy         and         c  :  d  :-,  ac  -,  ad, 
(504.  Rectangles  of  equal  altitudes  are  to  each  other  as  their  bases.) 


But,  by  hypothesis, 


a  '.  b  '.:  c  ',  d, 


therefore,  by  491, 


.*.     ac  :  be   : :  ae  :  ad; 


be  =  ad. 


198 


THE  ELEMENTS  OF  GEOMETRY. 


530.  Inverse.  If  two  rectangles  are  equivalent,  the  sides 
of  the  one  will  form  the  extremes,  and  the  sides  of  the  other 
the  means,  of  a  proportion. 


^■c 


a^. 


Hypothesis.     Reciangle  ad  =  be. 
Conclusion,     a  \  b  w  c  \  d. 

Proof.     Since  ad  =  be,  therefore,  by  489,  ac   :  be 
but  ae  :  be  ::  a  :  bj  and  ae  :  ad  '.'.  e  :  d, 

.'.     a     :    b    we    :  d. 


ae   :  ad\ 


531.  Corollary.     If  three  sects  are  proportional,  the  rect- 
angle of  the  extremes  is  equivalent  to  the  square  on  the  mean. 


Theorem   XIV. 

532'  If  two  chords  intersect  either  within  or  without  the 
circle,  the  rectangle  contained  by  the  segments  of  the  ofie  is  equiv- 
alent to  the  rectangle  contained  by  the  segments  of  the  other. 


RATIO  APPLIED. 


199 


Hypothesis.     Let  the  chords  AB  and  CD  intersect  in  P. 

Conclusion.     Rectangle  AP .  PB  =  rectangle  CP .  PD. 
Proof.     ^PAC  ==  ^  PDB, 

(376.  Angles  in  the  same  segment  of  a  circle  are  equal.) 


and 


^  APC  =  ^  BPD, 
.-.     A  APC  -  A  BPD, 
(508.  Equiangular  triangles  are  similar.) 

/.     AP  :   CP  ::PI>  :  PB, 
.-.     AP.PB  =  CP.PD, 


533.  Corollary.  Let  the  point  P  be  without  the  circle, 
and  suppose  DCP  to  revolve  about  P  until  C  and  D  coincide ; 
then  the  secant  DCP  becomes  a  tangent,  and  the  rectangle 


CP .  PD  becomes  the  square  on  PC.  Therefore,  if  from  a  point 
without  a  circle  a  secant  and  tangent  be  drawn,  the  rectangle 
of  the  whole  secant  and  part  outside  the  circle  is  equivalent  to 
the  square  of  the  tangent. 


200  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XV. 

534.  The  rectangle  of  two  sides  of  a  triangle  is  equivalent  to 
the  rectangle  of  two  sects  drawn  from  that  vertex  so  as  to  make 
equal  angles  with  the  two  sides ^  and  produced^  one  to  the  base, 
the  other  to  the  circle  circumscribing  the  triangle. 


Hypothesis.     ^  ABE  =  ^  CBD. 
Conclusion.     Rectangle  AB  .BC  —  DB .  BE, 
Proof.    Join  AE.    Then 

4-C^i.E, 
(376.  Angles  in  the  same  segment  of  a  circle  are  equal.) 

^  CBD  =  4.  ABE,  by  hypothesis, 

.-.     A  CBD  ~  A  ABEy 

.-.     AB  :  BE  ::  DB  '.  BC, 

.-.    AB.BC  ^  DB.BE. 

535-  Corollary  I.  If  BD  and  BE  coincide,  they  bisect 
the  angle  B ;  therefore  rectangle  AB  .  BC  ^  DB  .  BE  = 
DB{BD  +  DE)  =  BD^  +  BD  .  DE  =  BD^  +  CD .  DA  (by 
532).     Therefore,  when  the  bisector  of  an  angle  of  a  triangle 


RATIO  APPLIED. 


201 


meets  the  base,  the  rectangle  of  the  two  sides  is  equivalent  to 
the  rectangle  of  the  segments  of  the  base,  together  with  the 
square  of  the  bisector. 


536.   Corollary   II.     If  BD  be  a  perpendicular,  BE  is  a 
diameter,  for  angle  BAE  is  then  right ;  therefore  in  any  triangle 


-2>     C 


the  rectangle  of  two  sides  is  equivalent  to  the  rectangle  of  the 
diameter  of  the  circumscribed  circle  by  the  perpendicular  to 
the  base  from  the  vertex. 


Exercises.  %Z.  Prove  that  the  inverse  of  535  does  not 
hold  when  AB  —  BC, 

89.  Prove  535  when  it  is  an  exterior  angle  which  is  bi- 
sected. 


202  THE  ELEMENTS  OF  GEOMETRY, 


Theorem   XVI. 

537.  The  rectangle  of  the  diagonals  of  a  quadrilateral  in- 
scribed in  a  circle  is  equivalent  to  the  sum  of  ihe  two  rectangles 
of  its  opposite  sides. 


Hypothesis.    ABCD  an  inscribed  quadrilaieral. 

Conclusion.     Rectangle  A C .  BD  =^  AB  .  CD  ■\-  BC .  DA, 
Proof.     By  164,  make  ^  DAF  =  ^  BAG. 
To  each  add  ^i^^C. 
Then,  in  A  s  ^  CZ?  and  ABF, 

^DAC  =  ^BAF', 
also 

4.ACD  =  ^ABD, 
(376.  Angles  inscribed  in  the  same  segment  are  equal.) 

.-.     AACD'^  ^ABF, 
(508.  Equiangular  triangles  are  similar.) 

/.    AC  :  AB  ::   CD  i  BF; 
therefore,  by  529, 

AC  .BF=  AB.  CD, 

Again,  by  construction,  ^  DAF  =  ^  BAC, 


RATIO  APPLIED. 


203 


Moreover,  ^  ADF  =  ^  ACB, 

(376.  Angles  inscribed  in  the  same  segment  are  equal.) 

(508.  Equiangular  triangles  are  similar.) 

.-.     AC  :  AD  ::   CB  :  DF'y 
therefore,  by  529, 

AC  ,FD  =  AD.BC, 

.-.    AB  .CD  +  BC  ,DA  =  AC  .BF+  AC  .FD 

=  AC{BF  +  FD)  =  AC  . ^Z>. 


Problem   I. 
538.   To  alter  a  given  sect  in  a  given  ratio. 


^ 

3 

/                                              k 

^.^ . 

* 

Given,  Me  raf/b  as  ihat  of  sect  a  to  sect  b^  and  given  the 
sect  AB. 

Required,  to  find  a  sect  to  which  AB  shall  have  the  same  ratio  as 
a  to  b. 

Construction.     Make  any  angle  C, 

On  one  arm  cut  off  CD  =  AB.  On  the  other  arm  cut  off  CF  =  a, 
and  FG  =  b. 

Join  DFy  and  through  G  draw  GH  \\  DF, 

.-.     AB  :  DH  '.:  a  :  b. 


204  ^-^^  ELEMENTS  OF  GEOMETRY. 

539.  This  is  the  same  as  finding  a  fourth  proportional  to 
three  given  sects. 

To  find  a  third  proportional  to  a  and  by  make  AB  =  ^  in 
the  above  construction. 

540.  Every  alteration  of  a  magnitude  is  an  alteration  in 
some  ratio. 

Two  or  more  alterations  are  jointly  equivalent  to  some  one 
alteration,  and  then  this  single  alteration  which  produces  the 
joint  effect  of  two  is  said  to  be  compounded  of  those  two. 

The  composition  of  the  ratios  of  a\,Q  b  and  (7  to  Z>  is  per- 
formed by  assuming  F,  altering  it  into  G,  so  that  F  :  G  w  a  :  by 
then  altering  G  into  77,  so  that  G  :  H  '.:  C  :  D. 

The  joint  effect  turns  F  into  H)  and  the  ratio  oi  F  to  H  is 
the  ratio  compounded  of  the  two  ratios,  a  :  b  and  C  :  D. 

541.  A  ratio  arising  from  the  composition  of  two  equal 
ratios  is  called  the  Duplicate  Ratio  of  either. 


Theorem  XVII. 

542.  Mutually  equiangular  parallelograms  have  to  one  another 
the  ratio  which  is  compounded  of  the  ratios  of  their  sides. 


Hypothesis.    In  /=?  ACy  ^  BCD  =  ^  HCF  of  £y  CG. 
Conclusion.    /^ACi^yCG^  ratio  compounded  oi  DC  :  CF, 
and^C  :  CH, 


RATIO  APPLIED.  205 


Proof.     Place  the  /u^  so  that  HC  and  CB  are  in  one  line ;  then, 
by  log,  DC  and  CF  are  in  one  line.     Complete  the  £j  BF. 
Then 

CD  AC  :  C7BF  .i  DC  :   CF, 
and 

£yBF  '.  n7CG  \\  BC  '.  CH, 

(505.  Parallelograms  of  equal  altitude  are  as  their  bases.) 

.'.    c7  AC  has  to  z:7  CG  the  ratio  compounded   oi  DC    :    CF  and 
BC   :   CH, 

543.  Corollary  I.  Triangles  which  have  one  angle  of  the 
one  equal  or  supplemental  to  one  angle  of  the  other,  being 
halves  of  equiangular  parallelograms,  are  to  one  another  in  the 
ratio  compounded  of  the  ratios  of  the  sides  about  those  angles. 

544.  Corollary  II.  Since  all  rectangles  are  equiangular 
parallelograms,  therefore  the  ratio  compounded  of  two  ratios 
between  sects  is  the  same  as  the  ratio  of  the  rectangle  con- 
tained by  the  antecedents  to  the  rectangle  of  the  consequents. 

If  the  ratio  compounded  of   a  :  a\  and  d  :  d\  be  written 

—  .  — ,  this  corollary  proves  ^  .  —  =  -^ ;  and  the  composition 
ad  a     0       ab 

of  ratios  obeys  the  same  laws  as  the  multiplication  of  fractions. 

Thus  ^  .  ^  =^,  and  so  the  duplicate  ratio  of  two  sects  is  the 
00       b^ 

same  as  the  ratio  of  the  squares  on  those  sects. 

It  will  be  seen  hereafter  that  the  special  case  of  542,  when 

the  parallelograms  are  rectangles,  is  made  the  foundation  of  all 

mensuration  of  surfaces. 

Exercises.  90.  If  mutually  equiangular  parallelograms  are 
equivalent,  so  are  rectangles  with  the  same  sides. 

91.  Equivalent  parallelograms  having  the  same  sides  as 
equivalent  rectangles  are  mutually  equiangular. 


206  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XVIII. 

545.  Similar  triangles  are  to  one  another  as  the  squares  on 
their  corresponding  sides. 


Let  the  similar  triangles  ABC^  AHK,  be  placed  so  as  to  have  the 
sides  AB,  AC,  along  the  corresponding  sides  AH,  AK,  and  therefore 
BC  II  HK.    Join  CH, 

Since,  by  505,  triangles  of  equal  altitude  are  as  their  bases, 

.-.     A  ABC  :  A  AHC  w  AB  \  AH, 
and  A  AHC  :  A  AHK  ::  AC  :  AK  =  AB  :  AH,  by  hypothesis ; 

AB     AB       AB^ 


A  ABC  :  A  AHK  = 


AH    AH      AH^ 


Exercises.  92.  The  ratio  of  the  surfaces  of  two  similar 
triangles  is  the  square  of  the  ratio  of  similitude  of  the  tri- 
angles. 

93.  If  the  bisector  of  an  angle  of  a  triangle  also  bisect  a 
side,  the  triangle  is  isosceles. 

94.  In  every  quadrilateral  which  cannot  be  inscribed  in  a 
circle,  the  rectangle  contained  by  the  diagonals  is  less  than  the 
sum  of  the  two  rectangles  contained  by  the  opposite  sides. 

95.  The  rectangles  contained  by  any  twp  sides  of  triangles 
inscribed  in  equal  circles  are  proportional  to  the  perpendiculars 
on  the  third  sides. 

96.  Squares  are  to  one  another  in  the  duplicate  ratio  of 
their  sides. 


RATIO  APPLIED. 


207 


Theorem   XIX. 

546.    Similar  polygons  are  to  each  other  as  the  squares  on 
their  corresponding"  sides,  ^  -g- 


Hypothesis.    ABCD  and  A'B'Clf  two   similar  polygons,    of 
which  BC  and  B'  C  are  corresponding  sides. 

Conclusion.     ABCD  :  A'B'C'iy  : :  AB^  :  A'B'\ 
Proof.     By  516,  the  polygons  may  be  divided  into  similar  triangles. 
By  545,  any  pair  of  corresponding  triangles  are  as  the  squares  on 
corresponding  sides, 

t^ABD         BJy  /\BCD         BC^ 


AA'B'iy       B'D^        AB'Cjy       B'C'^ 


therefore,  by  496, 


A  ABD  +  A  BCD  BC 


A  A'B'D^  +  A  B'C'D^       B'C'^ 


In  the  same  way  for  a  third  pair  of  similar  triangles,  etc. 


208  THE  ELEMENTS  OF  GEOMETRY. 

547.  Corollary.  If  three  sects  form  a  proportion,  a  poly- 
gon on  the  first  is  to  a  similar  polygon  similarly  described  on 
the  second  as  the  first  sect  is  to  the  third. 


Theorem  XX. 

548.  The  perimeters  of  any  two  regular  polygons  of  the  same 
number  of  sides  have  the  same  ratio  as  the  radii  of  their  circum- 
scribed circles. 

7) 


Proof.  The  angles  of  two  regular  polygons  of  the  same  number 
of  sides  are  all  equal,  and  the  ratio  between  any  pair  of  sides  is  the 
same,  therefore  the  polygons  are  similar. 

But  lines  drawn  from  the  center  to  the  extremities  of  any  pair  of 
sides  are  radii  of  the  circumscribed  circles,  and  make  similar  triangles, 


therefore,  by  496, 


AB 

BC 

r 

"    A'B'~ 

B'C 

■?' 

AB  +  BC 

r 

CD 

A'B'  4-  B'C 

r' 

c'jy' 

AB  +  BC  -{-  CD 

r 

A'B'  ^B'C  ^  C'jy       r' 


RATIO   APPLIED.  209 


Problem   II. 
549.   To  find  a  mean  proportional  between  two  giveji  sects. 


3) 


Let  AB,  BC,  be  the  two  given  sects.  Place  AB,  BC,  in  the  same 
line,  and  on  AC  describe  the  semicircle  ADC.  From  B  draw  BD 
perpendicular  to  ^C 

BD  is  the  mean  proportional. 

(383.  An  angle  inscribed  in  a  semicircle  is  a  right  angle.) 

(521.  If  from  the  right  angle  a  perpendicular  be  drawn  to  the  hypothenuse,  it  will 

be  a  mean  proportional  between  the  segments  of  the  hypothenuse.) 

Exercises.  97.  If  the  given  sects  were  ^C  and  ^C  placed 
as  in  the  above  figure,  how  would  you  find  a  mean  proportional 
between  them  1 

98.  Half  the  sum  of  two  sects  is  greater  than  the  mean 
proportional  between  them. 

99.  The  rectangle  contained  by  two  sects  is  a  mean  propor- 
tional between  their  squares. 

100.  The  sum  of  perpendiculars  drawn  from  any  point 
within  an  equilateral  triangle  to  the  three  sides  equals  its 
altitude. 

loi.  The  bisector  of  an  angle  of  a  triangle  divides  the 
triangle  into  two  others,  which  are  proportional  to  the  sides 
of  the  bisected  angle. 

102.  Lines  which  trisect  a  side  of  a  triangle  do  not  trisect 
the  opposite  angle. 

103.  In  the  above  figure,  if  AF  ±  AD  meet  DB  produced  at 
F,  then  a  ABD  —  a  FOB. 


2IO 


THE  ELEMENTS  OF  GEOMETRY, 


Problem   III. 

550.  On  a  given  sect  to  describe  a  polygon  similar  to  a  given 
polygon. 


Let  ABCDE  be  the  given  polygon,  and  A'B'  the  given  sect. 

^OYCiAD,  AC, 

4.  A'B'C  =  4.  ABC, 
^B'A'C  =  4BAC,     ■ 
.-.     tB'C'A'  =  4BCA, 


Make 


and 
Then  make 


A  A'B'C 


h.ABC, 


and 


^.A'C'jy  =  4ACD, 

4.  C'A'jy  =  4  CAD, 

:,     4  CD  A'  =  4  CDA, 

t^AC'iy  ~  A^CA  etc. 

Therefore,  from  the  first  pair  of  similar  triangles, 

A'B'    :  B'C    ::  AB    :  BC, 
and 

B'C  :   CA'    ::  BC   :   CA. 

From  the  second  pair  of  similar  triangles, 

CA'  :   C'ly   ::   CA    :   CD, 
,',     B'C  :   CD'   ::  BC  :   CD, 
and  so  on. 

Thus  all  the  ^^s  in  one  polygon  are  =  the  corresponding  2<s  in  the 
other,  and  the  sides  about  the  corresponding  ^s  are  proportional. 
.*.     the  polygons  are  similar. 


RATIO  APPLIED. 


211 


Theorem   XXI. 

551.  In  a  right-angled  triangle,  any  polygon  upon  the  hypoth- 
enuse  is  equivalent  to  the  sum  of  the  similar  and  similarly 
described  polygons  on  the  other  two  sides. 


Draw  CD  ±  AB. 

Therefore,  by  520, 

AB  :  AC   ::  AC  :  AD', 
therefore,  by  547, 

AB  '.  AD  ::     Q    :  S. 
Similarly, 

AB  :  BD  ::     Q    :  R, 
Therefore,  by  496, 

AB  :  AD  -{-  DB  ::   Q  :  R  -{■  S, 

.',        Q=:    R   +   S. 

Exercises.  104.  If  551  applies  to  semicircles,  show  the 
triangle  equivalent  to  two  crescent-shaped  figures,  called  the 
lunes  of  Hippocrates  of  Chios  (about  450  B.C.). 


212  THE  ELEMENTS  OF  GEOMETRY. 


Problem  IV. 

552.   To  describe  a  polygon  equivalent  to  one^  and -similar  to 
anothery  given  polygon. 


JJC 


A. 


Let  D  be  the  one,  and  ABC  the  other,  given  polygon. 

By  262,  on  ^C  describe  any  ^7  CE  =  ABC,  and  on  AB  describe 
/:7  AM  =  Z>,  and  having  ^  AEM  =  ^  CAE. 

By  549,  between  AC  and  AF  find  a  mean  proportional  GH. 

^y  55°>  o^  ^^  construct  the  figure  KGH  similar  and  similarly 
described  to  the  figure  ABC. 

KGH  is  the  figure  required. 

It  may  be  proved,  as  in  542,  that  ^C  and  AF  are  in  one  line,  and 
also  LE  and  EM-, 
therefore,  by  505, 

AC  '.  AF  '.'.  n7  CE  '.  n7  AM  : :  ABC  :  £>, 
But 

AC  :  AF  ::  ABC  :  KGH, 

(547.  If  three  sects  form  a  proportion,  a  polygon  on  the  first  is  to  a  similar  polygon 
on  the  second  as  the  first  sect  is  to  the  third.) 

.-.    ABC  \    D     .'.  ABC  :  KGH-, 
therefore,  by  491, 

KGH  =  D, 


GEOMETRY  OF  THREE  DIMENSIONS. 


BOOK   VII. 


OF  PLANES  AND   LINES. 

553.  Already,  in  50,  a  plane  has  been  defined  as  the  surface 
generated  by  the  motion  of  a  line  always  passing  through  a 
fixed  point  while  it  slides  along  a  fixed  line. 

554.  Already,  in  97,  the  theorem  has  been  assumed,  that,  if 
two  points  of  a  line  are  in  a  plane,  the  whole  line  lies  in  that 
plane. 

Two  other  assumptions  will  now  be  made :  — 

555.  Any  number  of  planes  may  be  passed  through  any  line. 


z 


2 


^ 


556.  A  plane  may  be  revolved  on  any  line  lying  in  it. 


ax3 


214  THE  ELEMENTS  OF  GEOMETRY. 

Theorem   I. 

557.   Through  two  intersecting  lines y  one  plane y  and  only  one^ 
passes. 

a 


Hypothesis.     Two  lines  AB,  BC,  meeting  in  B. 

Conclusion.     One  plane,  and  only  one,  passes  through  them. 

Proof.  By  555,  let  any  plane  FG  be  passed  through  AB,  and,  by 
556,  be  revolved  around  on  AB  as  an  axis  until  it  meets  any  point  C 
of  the  line  BC, 

The  line  BC  then  has  two  points  in  the  plane  FG'j  and  therefore, 
by  554,  the  whole  line  BC'i^m  this  plane. 

AlsOy  any  plane  containing  AB  and  BC  must  coincide  with  FG. 

For  let  Q  be  any  point  in  a  plane  containing  AB  and  BC. 

Draw  QMN  in  this  plane  to  cut  AB,  BC,  inM  and  JV.  Then, 
since  M  and  JV  are  points  in  the  plane  FF,  therefore,  by  554,  Q  is  a 
point  in  the  plane  FF. 

Similarly,  any  point  in  a  plane  containing  AB,  BC,  must  lie  in  FF; 
therefore  any  plane  containing  AB,  BC,  must  coincide  with  FF. 

558-  Corollary  I.  Two  lines  which  intersect  lie  in  one 
plane,  and  a  plane  is  completely  determined  by  the  condition 
that  it  passes  through  two  intersecting  lines. 

559*  Corollary  II.  Any  number  of  lines,  each  of  which 
intersects  all  the  others  at  different  points,  lie  in  the  same 
plane ;  but  a  line  may  pass  through  the  intersection  of  two 
others  without  being  in  their  plane. 


OF  PLANES  AND  LINES.  21 5 

560.  Corollary  III.     A  line y  and  a  point  without  that  line, 
determine  a  plane, 

s 


Proof.    Suppose  AB  the  line,  and  C  the  point  without  AB. 

Draw  the  hne  CD  to  any  point  D  in  AB.  Then  one  plane  con- 
tains AB  and  CD,  therefore  one  plane  contains  AB  and  C. 

Again,  any  plane  containing  AB  must  contain  D ;  therefore,  any 
plane  containing  AB  and  C  must  contain  CD  also. 

But  there  is  only  one  plane  that  can  contain  AB  and  CD. 

Therefore  there  is  only  one  plane  that  can  contain  AB  and  C. 

Hence  the  plane  is  completely  determined. 

561.  Corollary  IV.  Three  points  not  in  the  same  line 
determine  a  plane. 


57 


For  let  A,  B,  C  be  three  such  points.     Draw  the  line  AB. 

Then  a  plane  which  contains  A,  B,  and  C  must  contain  AB 
and  C\  and  a  plane  which  contains  AB  and  C  must  contain  A, 
B,  and  C.  Now,  AB  and  C  are  contained  by  one  plane,  and  one 
only ;  .therefore  A,  B,  and  C  are  contained  by  one  plane,  and 
one  only.     Hence  the  plane  is  completely  determined. 


2l6  THE  ELEMENTS  OF  GEOMETRY. 

562.     Two  parallel  lines  determine  a  plane. 

For,  by  the  definition  of  parallel  lines,  the  two  lines  are  in 
the  same  plane ;  and,  as  only  one  plane  can  be  drawn  to  con- 
tain one  of  the  lines  and  any  point  in  the  other  line,  it  follows 
that  only  one  plane  can  be  drawn  to  contain  both  lines. 


Theorem   II. 

563.  If  two  planes  cut  one  another^  their  common  section  must 
be  a  straight  line, 

JL 


Hypothesis.     Lei  AB  and  CD  be  two  intersecting  planes. 

Conclusion.     Their  common  section  is  a  straight  line. 

Proof.  Let  M  and  N  be  two  points  common  to  both  planes. 
Draw  the  straight  line  MN.  Therefore,  by  554,  since  M  and  N  are  in 
both  planes,  the  straight  line  MN  lies  in  both  planes. 

And  no  point  out  of  this  line  can  be  in  both  planes ;  because  then 
two  planes  would  each  contain  the  same  line  and  the  same  point  without 
it,  which,  by  560,  is  impossible. 

Hence  every  point  in  the  common  section  of  the  planes  lies  in  the 
straight  line  MN. 


OF  PLANES  AND  LINES.  21/ 

564.  CoNTRANOMiNAL  OF  554.  A  line  which  does  not  lie 
altogether  in  a  plane  may  have  no  point,  and  cannot  have  more 
than  one  point,  in  common  with  the  plane. 

Therefore  three  planes  which  do  not  pass  through  the  same 
line  cannot  have  more  than  one  point  in  common ;  for,  by  563, 
the  points  common  to  two  planes  lie  on  a  line,  and  this  line  can 
have  only  one  point  in  common  with  the  third  plane. 

565.  All  planes  are  congruent ;  hence  properties  proved  for 
one  plane  hold  for  all.     A  plane  will  slide  upon  its  trace. 

Principle  of  Duality. 

566.  When  any  figure  is  given,  we  may  construct  a  reciprocal 
figure  by  taking  planes  instead  of  points,  and  points  instead  of 
planes,  but  lines  where  we  had  lines. 

The  figure  reciprocal  to  four  points  which  do  not  lie  in  a 
plane  will  consist  of  four  planes  which  do  not  meet  in  a  point. 
From  any  theorem  we  may  infer  a  reciprocal  theorem. 


Two  points  determine  a  line. 

Three  points  which  are  not  in 
a  line  determine  a  plane. 

A  line  and  a  point  without  it 
determine  a  plane. 

Two  lines  in  a  plane  determine 
a  point. 


Two  planes  determine  a  line. 

Three  planes  which  do  not  pass 
through  a  line  determine  a  point. 

A  line  and  a  plane  not  through 
it  determine  a  point. 

Two  lines  through  a  point  de- 
termine a  plane. 


There  is  also  a  more  special  principle  of  duality,  which,  in 
the  plane,  takes  points  and  lines  as  reciprocal  elements ;  for 
they  have  this  fundamental  property  in  common,  that  two  ele- 
ments of  one  kind  determine  one  of  the  other.  Thus,  from  a 
proposition  relating  to  lines  or  angles  in  axial  symmetry,  we  get 
a  proposition  relating  to  points  or  sects  in  central  symmetry. 


The  angle  between  two  corre- 
sponding lines  is  bisected  by  the 
axis. 


The  sect  between  tvvo  corre- 
sponding points  is  bisected  by  the 
center. 


2l8 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   III. 

567*  If  ^  ^^^^  ^^  perpendicular  to  two  lines  lying  in  a  plane y 
it  will  be  perpendicjilar  to  every  other  li7te  lying  in  the  plane  and 
passing  through  its  foot. 


Hypothesis.    Let  the  line  EF  be  perpendicular  to  each  of  the 
lines  AB,  CD,  at  E,  the  point  of  their  intersection. 

Conclusion.    EF  _L  GH,  any  other  line  lying  in  the  plane  ABCD, 
and  passing  through  E. 

Proof.    Take  AE  =  BE,  and  CE  =  DE. 

Join  AD  and  BC,  and  let  G  and  H  be  the  points  in  which  the 
joining  lines  intersect  the  third  line  GH. 

Take  any  point  F  in  EF.    Join  FA,  FB,  EC,  FD,  EG,  FH. 
Then 

A  AED  ^  A  BEC, 
(124.  Triangles  having  two  sides  and  the  included  angle  equal  are  congruent.) 

.-.    AD  =  BC,         and         ^  DAE  =  ^  CBE. 
Then 

A  AEG  ^  A  BEH, 
(128.  Triangles  having  two  angles  and  the  included  side  equal  are  congruent.) 

/.     GE  =  HE,        and  AG  =  BH. 


OF  PLANES  AND  LINES.  219 

Then 

A  AEF  ^  A  BEF, 

(124.  Triangles  having  two  sides  and  the  included  angle  equal  are  congruent.) 

AF  =  BF. 
In  the  same  way 

CF  =  DF, 
Then 

t.ADF^  t.BCF, 
(129.  Triangles  having  three  sides  in  each  respectively  equal  are  congruent.) 

/.     4.  DAF  =  4.  CBF, 
Then 

A  AFG  ^  A  BFH, 
(124.  Triangles  having  two  sides  and  the  included  angle  equal  are  congruent.) 

FG  =  FH, 

Then 

A  FEG  ^  A  FEH, 

(129.  Triangles  having  three  sides  in  each  respectively  equal  are  congruent.) 

/.     4  FEG  =  i.  FEB, 
FE  A.  GH. 

As  GH  is  any  line  whatever  lying  in  the  plane  ABCDy  and  passing 
through  E, 

/.    EF  J_  every  such  line. 

568.  A  line  meeting  a  plane  so  as  to  be  perpendicular  to 
every  line  lying  in  the  plane  and  passing  through  the  point  of 
intersection,  is  said  to  be  perpendicular  to  the  plane.  Then 
also  the  plane  is  said  to  be  perpendicular  to  the  line. 

569.  Corollary.  At  a  given  point  in  a  plane,  only  one 
perpendicular  to  the  plane  can  be  erected ;  and,  from  a  point 
without  a  plane,  only  one  perpendicular  can  be  drawn  to  the 
plane. 


220 


THE  ELEMENTS  OF  GEOMETRY. 


570.  Inverse  of  567.     All  lines  perpendicular  to  another 
line  at  the  same  point  lie  in  the  same  plane. 


Hypothesis.  AB  any  line  ±  BD  and  BE,  BC  any  other 
line  _L  AB. 

Conclusion.    ^C  is  in  the  plane  BDE, 

Proof.  For  if  not,  let  the  plane  passing  through  AB,  BC,  cut  the 
plane  BDE  in  the  line  BF.  Then  AB,  BC,  and  BE  are  all  in 
one  plane  ;  and  because  AB  J_  BD  and  BE,  therefore,  by  567, 

AB  J.  BF, 

But,  by  hypothesis,  AB  1.BC; 
therefore,  in  the  plane  ABE  we  have  two  lines  BC,  BF,  both  ±  AB  at 
B,  which,  by  105,  is  impossible ; 

.-.    BC  lies  in  the  plane  DBE. 


571.  Corollary.  If  a  right  angle  be  turned  round  one  of 
its  arms  as  an  axis,  the  other  arm  will  generate  a  plane ;  and 
when  this  second  arm  has  described  a  perigon,  it  will  have 
passed  through  every  point  of  this  plane. 

572.  Through  any  point  D  without  a  given  line  AB  to  pass 
a  plane  perpendicular  to  AB.  In  the  plane  determined  by 
the  line  AB  and  the  point  D,  draw  DB  J_  AB ;  then  revolve  the 
rt.  4.  ABD  about  AB. 


OF  PLANES  AND   LINES. 


221 


Theorem   IV. 

573.  Lines  perpendicular  to  the  same  plane  are  parallel  to 
each  other. 


Hypothesis.    AB,  CD  JL  plane  MN  at  the  points  B,  D. 

Conclusion.    AB  \\  CD. 

Proof.    Join  BD,  and  draw  DE  i.  BD  in  the  plane  MN, 
Make  DE  =  AB,  and  join  BE,  AD,  AE.    Then 
A  ABD  ^  A  EDB, 
(124.  Triangles  having  two  sides  and  the  included  angle  equal  are  congruent.) 

/.     AD  =  BE, 
Then 

A  ABE  ^  A  EDA, 

(129.  Triangles  having  three  sides  in  each  respectively  equal  are  congruent.) 

.-.     4.  ABE  =  4.  EDA, 
But,  by  hypothesis,  ABE  is  a  rt.  4, 

:.    EDA  is  a  rt.  ^, 

.'.    AD,  BD,  CD,  are  all  in  one  plane. 

(570.  All  lines  perpendicular  to  another  at  the  same  point  lie  in  the  same  plane.) 

But  AB  is  in  this  plane,  since  the  points  A  and  B  are  in  it, 
.'.    AB  and  CD  He  in  the  same  plane ; 
and  they  are  both  X  BD, 

,'.    AB  II   CD. 
(166.  If  two  interior  angles  are  supplemental,  the  lines  are  parallel.) 


222 


THE  ELEMENTS  OF  GEOMETRY. 


574.  Inverse  of  573.     If  one  of   two  parallels  is  perpen- 
dicular to  a  plane,  the  other  is  also  perpendicular  to  that  plane. 


7 


Hypothesis.    AB  \\  CD,     CD  ±  plane  MN. 
Conclusion.    AB  _L  plane  MN. 

Proof.     For  if  AB^  meeting  the  plane  MN  at  B^  is  not  ±  it,  let 
BF  \>Q  l.iX.'y 

therefore,  by  573,  BF  \\  CD,  and,  by  hypothesis,  BA  \\  CD, 
But  this  is  impossible, 

(99.  Two  intersecting  lines  cannot  both  be  parallel  to  the  same  line.) 
.-.    ^^  ±  plane  JfiV. 

575.  Corollary  I.  If  one  plane  be  perpendicular  to  one  of 
two  intersecting  lines,  and  a  second  plane  perpendicular  to  the 
second,  their  intersection  is 
perpendicular  to  the  plane  of 
the  two  lines.  - 

For  their  intersection  DF 
is  perpendicular  to  a  line 
through  D  parallel  to  ABy 
and  also  perpendicular  to  a 
line  through  D  parallel  to 
BC. 

(574.  If,  of  two  parallels,  one  be  perpendicular  to  a  plane,  the  other  is  also.) 

576.  Corollary  II.     Two  lines,  each  parallel  to  the  same 
line,  are  parallel  to  each  other,  even  though  the  three  be  not  in 


OF  PLANES  AND  LINES.  •  223 

one  plane.  For  a  plane  perpendicular  to  the  third  line  will,  by 
574,  be  perpendicular  to  each  of  the  others;  and  therefore, 
by  573,  they  are  parallel. 

Problem   I. 

577.   To  draw  a  line  perpendicular  to  a  given  plane  from  a 
given  point  without  it. 


GrvTEN,  ihe  plane  BH  and  the  point  A  without  it. 

Required,  to  draw  from  A  a  line  ±  plane  BH. 

Construction,     In  the  plane  draw  any  line  BC,  and,  by  139,  from 
A  drsLW  AD  ±BC. 

In  the  plane,  by  135,  draw  Z>jF  ±  BC;   and  from  A,  draw,  by  139, 
AF  ±  DF.    AF  will  be  the  required  perpendicular  to  the  plane. 

Proof.    Through  F  draw  GH  \\  BC. 

Then,  since  BC  1.  the  plane  ADFy 
(567.  A  line  perpendicular  to  any  two  lines  of  a  plane  at  their  intersection  is 
perpendicular  to  the  plane.) 

.\     GH  ±  the  plane  ADF, 
(574.  If  two  lines  are  parallel,  and  one  is  perpendicular  to  a  plane,  the  other  is 

also.) 

/.     GH  ±  the  line  AF  of  the  plane  ADF, 
:.    AF1.GH. 
But,  by  construction,  AF  _L  DF, 

,',    AF  J_  the  plane  passing  through  GH,  DF. 


224 


THE  ELEMENTS  OF  GEOMETRY. 


Problem   II. 

578.   To  erect  a  perpendicular  to  a  given  plane  from  a  given 
point  in  the  plane. 


1 


Let  A  be  the  given  point. 

From   any  point  B,  without  the  plane,  draw,  by  577,  -5  C  ±  the 
plane,  and  from  A,  by  167,  draw  AD  \\  BC, 

/.    AD  ±  the  plane. 
{574.  If  one  of  two  parallels  is  perpendicular  to  a  given  plane,  the  other  is  also.) 

579.  The  projection  of  a  point  upon  a  plane  is  the  foot  of 
the  perpendicular  drawn  from  the  point  to  the  plane. 


580.  The  projection  of  a  line  upon  a  plane  is  the  locus  of 
the  feet  of  the  perpendiculars  dropped  from  every  point  of  the 
line  upon  the  plane. 


OF  PLANES  AND   LINES. 


225 


These  perpendiculars  are  all  in  the  same  plane,  since  any 
two  are  parallel,  and  any  third  is  parallel  to  either,  and  has  a 
point  in  their  common  plane,  and  therefore  lies  wholly  in  that 
plane ;  therefore  the  projection  of  a  line  is  a  line. 


Theorem  V. 

581.  A  line  makes  with  its  own  projectiojt  upon  a  plane  a  less 
angle  than  with  any  other  line  ijt  the  plane. 


Hypothesis.  Lq^  BA  meet  the  plane  MN  at  B,  and  let  BA' 
be  its  project/on  upon  the  plane  MN^  and  BC  any  other  line  drawn 
through  B  in  the  plane  MN. 

Conclusion.     ^  ABA'  <  ^  ABC. 

Proof.     Take  BC  ^  BA.    Join  A C  and  A' C. 

Then,  in  As  ABA'  and  ABC, 

AB  =  AB, 

BA'  =  BC, 

AA'  <  AC, 

(150.  The  perpendicular  is  the  least  sect  from  a  point  to  a  line.) 

.-.     ^  ABA'  <^ABC. 
(161.  In  two  triangles,  ii  a  =  a',  6  =  y,  c  Kc*,  therefore  C<  C.) 


226  THE  ELEMENTS  OF  GEOMETRY, 

582.  The  angle  between  a  line  and  its  projection  on  a  plane 
is  called  the  Inclinatioji  of  the  line  to  the  plane. 

583.  Parallel  planes  are  such  as  never  meet,  how  far  soever 
they  may  be  produced. 

584.  A  line  is  parallel  to  a  plane  when  they  never  meet,  how 
far  soever  they  may  be  produced. 


Theorem  VI. 

585.  Planes  to  which  the  same  line  is  perpendicular  are  par- 
allel. 

For,  if  not,  they  intersect.  Call  any  point  of  their  intersection  X. 
Draw  from  X  a  line  in  each  plane  to  the  foot  of  the  common  perpen- 
dicular. Then  from  this  point  X  we  would  have  two  perpendiculars  to 
the  same  line,  which  is  impossible, 

(145.  There  can  be  only  one  perpendicular  from  a  point  to  a  line.) 

.*.     the  planes  cannot  intersect,  and 

.*.    are  parallel. 

586.  If  two  lines  are  parallel,  every  plane  through  one  of 
them,  except  the  plane  of  the  parallels,  is  parallel  to  the  other. 

Let  AB  and  GH  be  the  parallels,  and  DEF  any  plane 
through  GH\  then  the  line  AB  and  the  plane  DEF  are  par- 
allel. For  the  plane  of  the  parallels  ABHG  intersects  the 
plane  DEF  in  the  line  GH\  and,  if  AB  could  meet  the  plane 
DEFy  it  could  meet  it  only  in  some  point  of  GH\  but  AB  can- 
not meet  GH,  since  they  are  parallel  by  hypothesis.  Therefore 
AB  cannot  meet  the  plane  DEF. 

Applications,  (i)  Through  any  given  line  a  plane  can  be 
passed  parallel  to  any  other  given  line. 

(2)  Through  any  given  point  a  plane  can  be  passed  parallel 
to  any  two  given  lines  in  space. 


OF  PLANES  AND   LINES. 


227 


Theorem  VII. 

587-  If  ci  pair  of  ijttersecting  lines  be  parallel  to  another 
pair,  bnt  not  in  the  same  plane  with  them,  the  plane  of  the  first 
pair  is  parallel  to  the  plane  of  the  secojtd  pair. 


Hypothesis.    AB  \\  DE,  and  BC  \\  EF. 

Conclusion.     Plane  ABC  II  plane  DEF. 

Proof.     From  B  draw  BH  J_  plane  DEF,  meeting  it  in  H, 

Through  B  draw  GB  \\  DE,  and  BK  \\  EF, 

.-.     GB  II  AB,        and    '    BK  \\  BC, 

(576.  Two  lines  each  parallel  to  the  same  line  are  parallel  to  each  other.) 

/.     -2^  ABB  +  ^  BBG  =  St.  ^. 
But  because  BB  was  drawn  J_  plane  DEF, 
/.     ^BBG^xt.^. 


So  '^, 

and,  in  same  way, 


:^ABB=  rt.  ^; 


^  CBB  =  rt.  ^  ; 
.-.    BB  A_rg\^nQABC, 
.♦.     plane  ABC  \\  plane  DEF. 
(585.  Planes  to  which  the  same  line  is  perpendicular  are  parallel.) 


228 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  VIII. 

588.  If  a  pair  of  intersecting  lines  be  parallel  to  another  pair, 
any  angle  made  by  the  first  pair  is  equal  ,or  supplemental  to  any 
angle  made  by  the  second  pair. 


Hypothesis.    AB  \\  DE,  and  BC  II  EF. 
Conclusion.     2^  ABC  =  or  supplemental  to  ^  DEF, 
Proof.     Join  BE. 
Since  AB  \\  DE, 

.'.    AB,  BE,  and  ED,  are  in  one  plane. 

In  this  plane,  from  A  draw  a  line  I|  BE. 

It  must  meet  the  line  DE.     Call  the  intersection  point  D. 

.'.    ABED  is  a  ,C7 ; 
.-.    AB  =  DE,        and        AD  =  BE. 
In  same  way, 

BC  =  EF,         and         CF  =  BE, 

.-.     AD  =  CF. 

But  since,  by  construction,  AD  \\  BE,  and  CF  \\  BE,  therefore, 
hysi6,  AD  II   CF, 

/.    ACFD  isa,/=7, 

(216.  If  any  two  opposite  sides  of  a  quadrilateral  are  equal  and  parallel,  it  is  a 

parallelogram.) 


OF  PLANES  AND  LINES.  229 

.-.    AC  =  DF, 

,\     A  ABC  ^  A  £>£F, 

(129.  Triangles  with  three  sides  of  the  one  equal  to  three  of  the  other  are  con- 
gruent.) 

.-.     ^  ABC  =  ^  JDjEK 


Theorem   IX. 

589.   If  two  parallel  planes  be  cut  by  another  plane y  their 
common  sections  with  it  are  parallel. 


Call  the  parallel  planes  A  and  B,  and  the  third  plane  X. 

Then  the  lines  of  intersection  are  in  one  plane,  since  they  both  lie 
in  the  plane  X. 

Again,  because  one  of  these  lines  is  in  the  plane  A^  and  the  other 
in  the  parallel  plane  B,  they  can  never  meet. 

Therefore  the  two  lines  are  in  one  plane,  and  can  never  meet ;  that 
is,  they  aire  parallel. 

590.  Corollary.  Parallel  sects  included  between  two  par- 
allel planes  are  equal. 


230 


THE  ELEMENTS  OF  GEOME^TRY. 


Theorem   X. 

591.   Parallel  lines  intersecting  the  same  plane  are  equally 
inclined  to  it. 


Hypothesis. 


Conclusion. 


AB    II   CD, 
BB'  ±  plane  ACF, 
niy  A.  plane  ACF. 
^  BAB'  =  V  Dcjy, 


Proof.     ^  ABB'  =  or  supplemental  to  ^  CDL/, 

(588.  If  a  pair  of  intersecting  lines  be  parallel  to  another  pair,  any  angle  made  by 
the  first  pair  is  equal  or  supplemental  to  any  angle  made  by  the  second  pair.) 

But  2<  ABB'  cannot  be  supplemental  to  ^  CDL/ ,  since  each  is  acute, 
.-.     -4.  ABB'  =  4.  CDjy-, 
and  therefore  their  complements  are  equal. 

592.  Two  lines  not  in  the  same  plane  are  regarded  as  mak- 
ing with  one  another  the  angles  included  by  two  intersecting 
lines  drawn  parallel  respectively  to  them. 

We  know,  from  588,  that  these  angles  will  always  be  the 
same,  whatever  the  position  of  the  point  of  intersection. 


OF  PLANES  AND  LINES,  23 1 


Theorem   XL 

593*  V  ^^^  //«^J  be  cut  by  three  parallel  planes,  the  corre- 
spondi7ig  sects  are  proporiio7ial. 


Let  the  lines  AB,  CD,  be  cut  by  the  ||  planes  MN,  FQ,  RSy  in  the 
points  Aj  E,  By  and  C,  F,  D. 

Conclusion.    AE  :  EB  ::   CF  :  FD. 
Join  AD,  cutting  the  plane  FQ  in  G, 
Join  A C,  BD,  EG,  EG.    Then 

EG  II  BD. 
(589.  If  parallel  planes  be  cut  by  a  third  plane,  their  common  sections  with  it  are 

parallel.) 

In  the  same  way,  AC  \\  GF\ 

,\    AE  :  EB   ::  AG  :   GD, 
and 

AG  '.   GD  ::   CF  ',  FD', 
(499.  A  line  parallel  to  the  base  of  a  triangle  divides  the  sides  proportionally.) 

/.    AE  \  EB   \\  CF  '.  FD. 


232  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XII. 

594.   Two  lines  not  in  tJu  same  plane  have  one,  and  only  oite^ 
common  j)erpendicular. 


3L         T 


Hypothesis.  AB  and  CD,  two  lines  not  in  tha  same  plane,  and 
therefore  neither  parallel  nor  intersecting  each  other. 

Conclusion.  There  is  one  line,  and  no  more,  perpendicular  to  both 
AB  and  CD. 

Proof.  Through  one  of  the  lines,  as  CD,  pass  a  plane,  and  let  it 
revolve  on  CD  as  axis  until  it  is  parallel  to  AB.  Call  this  plane  MN. 
Let  A'B'  be  the  projection  of  AB  on  the  plane  MN,  and  let  O  be  the 
point  in  which  this  projection  intersects  CD.  Then  O,  like  every  point 
in  the  projection,  is  the  foot  of  a  perpendicular  to  the  plane  from  some 
point  of  AB.  Call  this  point  P.  Then,  since  FO  is  perpendicular  to 
the  plane  MN,  it  is  perpendicular  to  CD  and  A'B'.  But  A'B'  is  par- 
allel to  AB,  because,  being  in  a  plane  parallel  to  AB,  it  can  never  meet 
AB ;  and,  being  the  projection  of  AB,  it  is,  by  580,  in  the  same  plane 
with  AB. 

.'.    -since  FO  _L  A'B',  it  is  also  ±  AB, 

(170.  A  line  perpendicular  to  one  of  two  parallels  is  perpendicular  to  the  other 

also.) 

,\     OF  ±  both  AB  and  CD. 
If  there  could  be  any  other  common  perpendicular,  call  it  F'Q. 


OF  PLANES  AND   LINES, 


233 


Through  Q  draw  in  the  plane  MN,  QR  \\  AB. 

Since  FQ  ±  AB,        .'.    FQ  J_  QR-, 
but,  by  hypothesis,  F  Q  1.  CD, 

:.    i^'Q  ±  plane  J/W; 

/.     ^  is  a  point  in  A'B' ,  the  projection  of  AB, 

.'.     Q  is  O,  the  only  point  common  to  A'B'  and  CD, 

.-.     jP' ^  coincides  with /'C?. 

(569.  At  a  given  point  in  a  given  plane,  only  one  perpendicular  to  the  plane  can  be 

erected.) 


Theorem    XIII. 

595.  The  smallest  sect  between  two  lines  not  i7t  the  same 
plane  is  their  common  perpeitdictdar. 

For  if  any  sect  drawn  from  one  to  the  other  is  not  perpendicular  to 
both,  by  dropping  a  perpendicular  to  the  hne  it  cuts  obliquely  from  the 
point  where  it  meets  the  other  line,  we  get  a  smaller  sect. 

(150.  The  perpendicular  is  the  smallest  sect  from  a  point  to  a  line.) 

596.  Remark.  If  two  planes  intersect,  and  two  intersecting 
lines  are  drawn,  one  in  each  plane,  these  lines  may,  for  the  same 


two  planes,  make  an  acute,  right,  or  obtuse  angle,  according  to 
their  relation  to  the  line  of  intersection  of  the  planes. 


234 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XIV. 

597.   The  smallest  sect  from  a  point  to  a  plane  is  the  perpen- 
dicidar. 


Hypothesis.    AD  ±  plane  MN.    B  any  other  point  of  MN. 
Conclusion.    AD  <  AB. 

Proof.     Join  AB,  BD.     In  A  ABDyX  ^^^  is  a  rt.  ^, 
.-.     AD  <  AB. 
(150.  The  perpendicular  is  the  least  sect  from  a  point  to  a  line.) 


Theorem   XV. 

598.  Equal  obliques  from  a  point  to  a  plane  m,eet  the  pla7te 
in  a  circle  whose  center  is  the  foot  of  the  perpendicular  from  the 
point  to  the  plane. 


OF  PLANES  AND  LINES. 


235 


Hypothesis.  AB,  AC,  equal  sects  from  A  to  the  plane  MN. 
AD  ±  plane  MN, 

Conclusion.    DB  =  DC. 

Proof.     Rt.  A  ABD  ^  rt.  A  A  CD. 

(179.  If  two  right  triangles  have  the  hypothenuse  and  one  side  respectively  equal, 

they  are  congruent.) 

599.  Corollary  I.  To  draw  a  perpendicular  from  a  point 
to  a  plane,  draw  any  oblique  from  the  point  to  the  plane ; 
revolve  this  sect  about  the  point,  tracing  a  circle  on  the  plane ; 
find  the  center  of  this  circle,  and  join  it  to  the  point. 

600.  Corollary  II.  If  through  the  center  of  a  circle  a 
line  be  passed  perpendicular  to  its  plane,  the  sects  from  any 
point  of  this  line  to  points  on  the  circle  are  equal. 


Theorem   XVI. 

601.  The  locus  of  all  points  from  which  the  two  sects  drawn 
to  two  fixed poiftts  are  equal,  is  the  pla7ie  bisecting  at  right  angles 
the  sect  joining  the  two  given  points. 


X 

For  the  line  from  any  such  point  to  the  mid  point  of  the  joining 
sect  is  perpendicular  to  that  sect ;  therefore  all  such  lines  form  a  plane 
perpendicular  to  that  sect  at  its  mid  point. 
(570.  All  lines  perpendicular  to  another  line  at  the  same  point  lie  in  the  same  plane.) 


2^6 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XVII. 

602.  The  locus  of  all  points  from  which  the  two  perpendicu- 
lars onto  the  same  sides  of  two  fixed  planes  are  equal,  is  a  plane 
determined  by  one  such  point  and  the  intersection  line  of  the  tzvo 
given  planes. 


Hypothesis.  Lei  AB  and  CD  be  the  two  given  planes,  and  K 
one  such  point. 

Conclusion.  Perpendiculars  dropped  on  to  AB  and  CD  from  any 
point  in  the  plane  determined  by  X  and  the  intersection  line  BC  are 
equal. 

Proof.  Take  P  any  point  in  the  plane  XCB.  Draw  PH  ±  plane 
AB,  and  BP  _L  plane  CB>.  Call  G  the  point  where  the  plane  FBI/ 
cuts  the  line  BC.     Join  FG,  BG,  GH. 

Draw  KC  ||  BG.  Because  the  perpendiculars  from  K  are  given 
equal,  therefore  KC  is  equally  inclined  to  the  two  planes.  But  BG  has 
the  same  inclination  to  each  as  KC ; 

(591.  Parallels  intersecting  the  same  plane  are  equally  inclined  to  it.) 

/.     ^  BGF  =  ^  BGH, 
/.     A  BGF  ^  A  BGH, 
BF     =  BH, 


OF  PLANES  AND   LINES.  237 


Theorem   XVIII. 

603 •  If  three  lines  not  in  the  same  plane  meet  at  one  pointy 
any  two  of  the  angles  formed  are  together  greater  than  the  third. 


Proof.  The  theorem  requires  proof  only  when  the  third  angle 
considered  is  greater  than  each  of  the  others.  In  the  plane  of  the 
greatest  4.  BAC,  make  4.  BAF  =  4  BAD.  Make  AF  =  AD. 
Through  F  draw  a  line  BFC  cutting  AB  in  B^  and  AC  m  C.     Join 

DB  and  DC. 

A  BAD  ^  A  BAF, 

(124.  Triangles  having  two  sides  and  the  included  angle  respectively  equal  are 

congruent.) 

.-.     BD  =  BF. 
But  from  A  BCD  we  have  BD  +  DC>  BC. 

{156.  Any  two  sides  of  a  triangle  are  together  greater  than  the  third.) 

And,  taking  away  the  equals  BD  and  BF, 
DC  >  FC, 
.-.     in  As  CAD  and  CAF,  we  have  4  CAD  >  4  CAR 

(161.  If  two  triangles  have  two  sides  respectively  equal,  but  the  third  side  greater 
in  the  first,  its  opposite  angle  is  greater  in  the  first.) 

Adding  the  equal  4  s  BAD  and  BAF  gives 

4  BAD  +  4  CAD  >  4  BAC. 


238  THR  ELEMENTS  OF  GEOMETRY. 


Theokem   XIX. 

604.  If  the  vertices  of  a  convex  polygon  be  joined  to  a  point 
not  in  its  plane ^  the  snin  of  the  vertical  angles  of  the  triangles  so 
made  is  less  than  a  perigon. 

s 


Proof.  The  sum  of  the  angles  of  the  triangles  which  have  the 
common  vertex  S  is  equal  to  the  sum  of  the  angles  of  the  same  number 
of  triangles  having  their  vertices  2X   O  in  the  plane  of  the  polygon. 

But 

%.  SAB  +  ^  SAE  >  ^  BAE, 

%.  SBA  4-  ^  SBC  >  t  ABC,  etc. 

(603.  If  three  Hnes  not  in  a  plane  meet  at  a  point,  any  two  of  the  angles  formed 
are  together  greater  than  the  third.) 

Hence,  summing  all  these  inequalities,  the  sum  of  the  angles  at  the 
bases  of  the  triangles  whose  vertex  is  S,  is  greater  than  the  sum  of 
the  angles  at  the  bases  of  the  triangles  whose  vertex  is  O  \  therefore 
the  sum  of  the  angles  at  ^  is  less  than  the  sum  of  the  angles  at  O,  that 
is,  less  than  a  perigon. 


BOOK   VIII. 


TRI-DIMENSIONAL  SPHERICS. 

605.  If  one  end  point  of  a  sect  is  fixed,  the  locus  of  the 
other  end  point  is  a  Sphere. 

606.  The  fixed  end  point  is  called  the  Center  of  the  sphere. 


607.  The  moving  sect  in  any  position  is  called  the  Radius  of 
the  sphere. 

608.  As  the  motion  of  a  sect  does  not  change  it,  all  radii  are 
equal. 

609.  The  sphere  is  a  closed  surface  ;  for  it  has  two  points  on 
every  line  passing  through  the  center,  and  the  center  is  midway 
between  them. 

•39 


240  THE  ELEMENTS  OF  GEOMETRY. 

6io.  Two  such  points  are  called  Opposite  Points  of  the 
sphere,  and  the  sect  between  them  is  called  a  Diameter. 

6ii.  The  sect  from  a  point  to  the  center  is  less  than,  equal 
to,  or  greater  than,  the  radius,  according  as  the  point  is  within, 
on,  or  without,  the  sphere. 

For,  if  a  point  is  on  the  sphere,  the  sect  drawn  to  it  from 
the  center  is  a  radius  ;  if  the  point  is  within  the  sphere,  it  lies 
on  some  radius  ;  if  without,  it  lies  on  the  extension  of  some 
radius. 

612.  By  33,  Rule  of  Inversion,  a  point  is  within,  on,  or  with- 
out, the  sphere,  according  as  the  sect  to  it  from  the  center  is 
less  than,  equal  to,  or  greater  than,  the  radius. 


Theorem   I. 
613.   The  common  section  of  a  sphere  and  a  plane  is  a  ciirle. 


Take  any  sphere  with  center  O. 

Let  A,  B,  C,  etc.,  be  points  common  to  the  sphere  and  a  plane, 
and  OD  the  perpendicular  from  O  to  the  plane.     Then 

OA  =  OB  =  OC  =^  etc., 
being  radii  of  the  sphere, 

.*.     ABC,  etc.,  is  a  circle  with  center  D. 

(598.  Equal  obliques  from  a  point  to  a  plane  meet  the  plane  in  a  circle  whose  center 
is  the  foot  of  the  perpendicular  from  the  point  to  the  plane.) 


TRI-DIMENSIONAL  SPHERICS.  24 1 

614.  Corollary.  The  line  through  the  center  of  any  circle 
of  a  sphere,  perpendicular  to  its  plane,  passes  through  the  cen- 
ter of  the  sphere. 

615.  A  Great  Circle  of  a  sphere  is  any  section  of  the  sphere 
made  by  a  plane  which  passes  through  the  center. 

All  other  circles  on  the  sphere  are  called  Small  Circles. 

616.  Corollary.  All  great  circles  of  the  sphere  are  equal, 
since  each  has  for  its  radius  the  radius  of  the  sphere. 

617.  The  two  points  in  which  a  perpendicular  to  its  plane, 
through  the  center  of  a  great  or  small  circle  of  the  sphere, 
intersects  the  sphere,  are  called  the  Poles  of  that  circle. 

618.  Corollary.  Since  the  perpendicular  passes  through 
the  center  of  the  sphere,  the  two  poles  of  any  circle  are  oppo- 
site points,  and  the  diameter  between  them  is  called  the  Axis 
of  that  circle. 


Theorem   II. 

619.  Every  great  circle  divides  the  sphere  into  two  congruent 
hemispheres. 


For  if  one  hemisphere  be  turned  about  the  fixed  center  of  the 
sphere  so  that  its  plane  returns  to  its  former  position,  but  inverted, 
the  great  circle  will  coincide  with  its  own  trace,  and  the  two  hemispheres 
will  coincide. 


242 


THE  ELEMENTS  OF  GEOMETRY. 


620.  Any  two  great  circles  of  a  sphere  bisect  each  other. 


Since  the  planes  of  these  circles  both  pass  through  the 
center  of  the  sphere,  their  line  of  intersection  is  a  diameter  of 
the  sphere,  and  therefore  of  each  circle. 

621.  If  any  number  of  great  circles  pass  through  a  point, 
they  will  also  pass  through  the  opposite  point. 


622.  Through  any  two  points  in  a  sphere,  not  the  extremi- 


ties of  a  diameter,  one,  and  only  one,  great  circle  can  be  passed  ; 
for  the  two  given  points  and  the  center  of  the  sphere  determine 


TRI-DIMENSIONAL  SPHERICS.  243 

its  plane.     Through  opposite  points,  an  indefinite  number  of 
great  circles  can  be  passed. 

623.  Through  any  three  points  in  a  sphere,  a  plane  can  be 
passed,  and  but  one ;  therefore  three  points  in  a  sphere  deter- 
mine a  circle  of  the  sphere. 

624.  A  small  circle  is  the  less  the  greater  the  sect  from  its 
center  to  the  center  of  the  sphere.  For,  with  the  same  hypoth- 
enuse,  one  side  of  a  right-angled  triangle  decreases  as  the  other 
increases.  ^««.^ 


625.  A  Zone  is  a  portion  of  a  sphere  included  between  two 
parallel  planes.  The  circles  made  by  the  parallel  planes  are 
the  Bases  of  the  zone. 

626.  A  line  or  plane  is  tangent  to  a  sphere  when  it  has  one 
point,  and  only  one,  in  common  with  the  sphere. 

627.  Two  spheres  are  tangent  to  each  other  when  they  have 
one  point,  and  only  one,  in  common. 

Exercises.  105.  If  through  a  fixed  point,  within  or  with- 
out a  sphere,  three  lines  are  drawn  perpendicular  to  each  other, 
intersecting  the  sphere,  the  sum  of  the  squares  of  the  three 
intercepted  chords  is  constant.  Also  the  sum  of  the  squares 
of  the  six  segments  of  these  chords  is  constant. 

106.  If  three  radii  of  a  sphere,  perpendicular  to  each  other, 
are  projected  upon  any  plane,  the  sum  of  the  squares  of  the 
three  projections  is  equal  to  twice  the  square  of  the  radius  of 
the  sphere. 


244  ^-^^  ELEMENTS  OF  GEOMETRY. 


Theorem   III. 

628.  A  plane  perpendicular  to  a  radius  of  a  sphere  at  its 
extremity  is  tangent  to  the  sphere. 


For,  by  597,  this  radius,  being  perpendicular  to  the  plane,  is  the 
smallest  sect  from  the  center  to  the  plane ;  therefore  every  point  of 
the  plane  is  without  the  sphere  except  the  foot  of  this  radius. 

629.  Corollary.  Every  line  perpendicular  to  a  radius  at 
its  extremity  is  tangent  to  the  sphere. 

630.  Inverse  of  628.  Every  plane  or  line  tangent  to  the 
sphere  is  perpendicular  to  the  radius  drawn  to  the  point  of 
contact.  For  since  every  point  of  the  plane  or  line,  except 
the  point  of  contact,  is  without  the  sphere,  the  radius  drawn 
to  the  point  of  contact  is  the  smallest  sect  from  the  center  of 
the  sphere  to  the  plane  or  line  ;  therefore,  by  597,  it  is  per- 
pendicular. 

Exercises.  107.  Cut  a  given  sphere  by  a  plane  passing 
through  a  given  line,  so  that  the  section  shall  have  a  given 
radius. 

108.  Find  the  locus  of  points  whose  sect  from  point  A  is  ^, 
and  from  point  B  \s  b. 


TRI-DIMENSIONAL   SPHERICS. 


245 


Theorem   IV. 

631.  If  two  spheres  cut  one  another ^  their  miersection  is  a  cir- 
cle whose  plane  is  perpendicnlar  to  the  line  Joining  the  centers  of 
the  spheres^  and  whose  center  is  in  that  line. 


Hypothesis.  Let  C  and  O  be  the  centers  of  the  spheres,  A 
and  B  any  two  points  in  their  intersection. 

Conclusion.  A  and  B  are  on  a  circle  having  its  center  on  the  line 
OC,  and  its  plane  perpendicular  to  that  line. 

Proof.    Join  CA,  CB,  OA,  OB.     Then 
A  CAO^  A  CBO, 
because  they  have   OC  common,  CA  =  CB,  and  OA  =  OBj  radii  of 
the  same  sphere. 

Since  these  As  are  =,  therefore  perpendiculars  from  A  and  B  upon 
OC  are  equal,  and  meet  OC  at  the  same  point,  Z>. 

Then  AB>  and  DB  are  in  a  plane  ±  OC ;  and,  being  equal  sects, 
their  extremities  A  and  B  axe  in  a  circle  having  its  center  at  Z>. 

632.  Corollary.  By  moving  the  centers  of  the  two  inter- 
secting spheres  toward  or  away  from  each  other,  we  can  make 
their  circle  of  intersection  decrease  indefinitely  toward  its 
center ;  therefore,  if  two  spheres  are  tangent,  either  internally 
or  externally,  their  centers  and  point  of  contact  lie  in  the  same 
line. 


246  THE  ELEMENTS  OF  GEOMETRY. 


Theorem  V. 

633.    Through  any  four  pohits  not  in  the  same  plane,  one 
spJiere,  and  only  one,  can  be  passed. 


Let  A^  B,  C,  D,  be  the  four  points.  Join  them  by  any  three  sects, 
as  AB,  BC,  CD.     Bisect  each  at  right  angles  by  a  plane. 

The  plane  bisecting  BC  has  the  line  EH'm.  common  with  the  plane 
bisecting  AB,  and  has  the  line  FO  in  common  with  the  plane  bisecting 
CD.     Moreover,  EH  i.  plane  ABC,  and  FO  ±  plane  BCD, 

(575.  If  one  plane  be  perpendicular  to  one  of  two  intersecting  lines,  and  a  second 
plane  perpendicular  to  the  second,  their  intersection  is  perpendicular  to  the 
plane  of  the  two  lines.) 

.-.     EH  LEG,        and        FO  JL  FG, 

,'.     EH  and  FO  meet,  since,  by  hypothesis,  ^  at  6^  >  o  and  <  st.  ^. 

Call  their  point  of  meeting  O.  O  shall  be  the  center  of  the  sphere 
containing  A,  B,  C,  and  D, 


TRI-DIMENSIONAL  SPHERICS. 


247 


For  O  is  in  three  planes  bisecting  at  right  angles  the  sects  AB,  BC, 
CD, 

(601.  The  locus  of  all  points  from  which  the  two  sects  drawn  to  two  fixed  points 
are  equal,  is  the  plane  bisecting  at  right  angles  the  sect  joining  the  two  given 
points.) 

634.  A  Tetrahedron  is  a  solid  bounded  by  four  triangular 
plane  surfaces  called  its  faces.  The  sides  of  the  triangles  are 
called  its  edges. 


635.  Corollary  I.  A  sphere  may  be  circumscribed  about 
any  tetrahedron. 

636.  Corollary  II.  The  lines  perpendicular  to  the  faces 
of  a  tetrahedron  through  their  circumcenters,  intersect  at  a 
common  point. 

637.  Corollary  III.  The  six  planes  which  bisect  at  right 
angles  the  six  edges  of  a  tetrahedron  all  pass  through  a  com- 
mon point. 


248  THE  ELEMENTS  OF  GEOMETRY. 

Problem   I. 
638.  To  inscribe  a  sphere  in  a  given  tetrahedron. 


Construction.  Through  any  edge  and  any  point  from  which  per- 
pendiculars to  its  two  faces  are  equal,  pass  a  plane. 

Do  the  same  with  two  other  edges  in  the  same  face  as  the  first,  and 
let  the  three  planes  so  determined  intersect  at  O.  O  shall  be  the  center 
of  the  required  sphere. 

(602.  The  locus  of  all  points  from  which  the  perpendiculars  on  the  same  sides  of 
two  planes  are  equal,  is  a  plane  determined  by  one  such  point  and  the  intersec- 
tion line  of  the  given  planes.) 

639.  Corollary.  In  the  same  way,  four  spheres  may  be 
escribed,  each  touching  a  face  of  the  tetrahedron  externally, 
and  the  other  three  faces  produced. 

640.  The  solid  bounded  by  a  sphere  is  called  a  Globe. 

641.  A  globe-segment  is  a  portion  of  a  globe  included  be- 
tween two  parallel  planes. 

The  sections  of  the  globe  made  by  the  parallel  planes  are 
the  bases  of  the  segment. 

Any  sect  perpendicular  to,  and  terminated  by,  the  bases,  is 
the  altitude  of  the  segment. 

642.  A  figure  with  two  points  fixed  can  still  be  moved  by 
revolving  it  about  the  line  determined  by  the  two  points. 


TR /-DIMENSIONAL  SPHERICS.  249 

This  revolution  can  be  performed  in  either  of  two  senses, 
and  continued  until  the  figure  returns  to  its  original  position. 
The  fixed  line  is  called  the  Axis  of  Revohction,  If  the  axis  of 
revolution  is  any  line  passing  through  the  center,  a  sphere 
slides  upon  its  trace.  This  is  because  every  section  of  a  sphere 
by  a  plane  is  a  circle. 

Any  figure  has  central  symmetry  if  it  has  a  center  which 
bisects  all  sects  through  it  terminated  by  the  surface.  The 
sphere  has  central  symmetry,  and  coincides  with  its  trace 
throughout  any  motion  during  which  the  center  remains  fixed. 
Thus  any  figure  drawn  on  a  sphere  may  be  moved  about  on  the 
sphere  without  deformation.  But,  unlike  planes,  all  spheres 
are  not  congruent.     Only  those  with  equal  radii  will  coincide. 

In  general,  a  figure  drawn  upon  one  sphere  will  not  fit  upon 
another.  So  we  cannot  apply  the  test  of  superposition,  except 
on  the  same  sphere  or  spheres  whose  radii  are  equal.  Again, 
if  we  wish  the  angles  of  a  figure  on  a  sphere  to  remain  the 
same  while  the  sides  increase,  we  must  magnify  the  whole 
sphere :   on  the  same  sphere  similar  figures  cannot  exist. 

643.  Two  points  are  symmetrical  with  respect  to  a  plane 
when  this  plane  bisects  at  right  angles  the  sect  joining  them. 
Two  figures  are  symmetrical  with  respect  to  a  plane  when 
every  point  of  one  figure  has  its  symmetrical  point  in  the  other. 
Any  figure  has  planar  symmetry  if  it  can  be  divided  by  a  plane 
into  two  figures  symmetrical  with  respect  to  that  plane.  The 
sphere  is  symmetrical  with  respect  to  every  plane  through  its 
center.  Any  two  spheres  are  symmetrical  with  respect  to 
every  plane  through  their  line  of  centers. 

Every  such  plane  cuts  the  spheres  in  two  great  circles ;  and 
the  five  relations  between  the  center-sect,  radii,  and  relative 
position  of  these  circles  given  in  410,  with  their  inverses,  hold 
for  the  two  spheres. 

Any  three  spheres  are  symmetrical  with  respect  to  the 
plane  determined  by  their  centers. 


250 


THE  ELEMENTS  OF  GEOMETRY, 


Theorem  VI. 

644.  If  a  sphere  he  tangent  to  the  parallel  plafzes  containing 
opposite  edges  of  a  tetrahedron^  and  sections  made  in  the  globe 
and  tetrahedron  by  one  plane  parallel  to  these  are  equivalenty 
sections  made  by  any  parallel  plaiie  are  equivalent. 


K 


K 


\^. ij Jo 

L\              /^  ^^ 

Hypothesis.     Let  KJ  be  the  sect  J_  the  edges  EF  and  GH  in 
the  II  tangent  planes. 

Then  Kf  =  DT,  the  diameter. 

Let  sections  made  by  plane  ±  Kf  at  R  and  ±  DT  at  I,  where 
KR  =  DI,  be  equivalent;  that  is,  ^  MO  =  O  PQ. 

Draw  any  parallel  plane  ACBLSN, 

Conclusion,    ^l^  LN  —  O  AB. 

Proof.     Since  A  LEU ^  h.  MEW,  and  A  LHV ^  A  MHZ, 
:.     MW  \  LU  w  EM  -.  EL  ::  fR    :  fS; 
MZ     '.  LV   -.'.  HM  '.  HL  '.:  KR  :  KS; 

(593.  If  lines  be  cut  by  three  parallel  planes,  the  corresponding  sects  are  propor- 
tional.) 

.-.     WM.  MZ  :   C/L.LF  ::  fR  .RK   :  fS  .  SK ; 
:,    ^MO  :  ^  LN  ::  fR  .RK  :  fS  .SK  (i) 

{542.  Mutually  equiangular  parallelograms  have  to  one  another  the  ratio  which  is 
compounded  of  the  ratios  of  their  sides.) 


TRI-DIMENSIONAL   SPHERICS. 


251 


But 

OPQ  \  Q)AB  '.'.  PP  '.  AC-   ::  TI .  ID  :  TC .  CD,     (2) 

{546.    Similar  figures  are  to  each  other  as  the  squares  on  their  corresponding 

sects.) 

(522.  If  from  any  point  in  a  circle  a  perpendicular  be  dropped  upon  a  diameter,  it 

will  be  a  mean  proportional  between  the  segments  of  the  diameter.) 

By  hypothesis  and  construction,  in  proportions  (i)  and  (2),  the 
first,  third,  and  fourth  terms  are  respectively  equal, 


Theorem  VII. 

645.   The  sects  Joining  its  pole  to  points  on  any  circle  of  a 
sphere  are  equal. 


Proof.  (600.  If  through  the  center  of  a  circle  a  line  be  passed 
perpendicular  to  its  plane,  the  sects  from  any  point  of  this  line  to  points 
on  the  circle  are  equal.) 

646.  Corollary.  Since  chord  PA  equals  chord  PB,  there- 
fore the  arc  subtended  by  chord  PA  in  the  great  circle  PA 
equals  the  arc  subtended  by  chord  PB  in  the  great  circle  PB. 

Hence  the  great-circle-arcs  joining  a  pole  to  points  on  its 
circle  are  equal. 


252 


THE  ELEMENTS  OF  GEOMETRY. 


So,  if  an  arc  of  a  great  circle  be  revolved  in  a  sphere  about 
one  of  its  extremities,  its  other  extremity  will  describe  a  circle 
of  the  sphere. 

647.  One-fourth  a  great  circle  is  called  a  Quadrant. 

648.  The  great-circle-arc  joining  any  point  in  a  great  circle 
with  its  pole  is  a  quadrant. 


p 

^ 

o\ 

"'^N 

/  'N 

\3j 

:/ 

::^ 

649.  If  a  point  P  be  a  quadrant  from  two  points,  A^  B,  which 
are  not  opposite,  it  is  the  pole  of  the  great  circle  through  A,  B ; 
for  each  of  the  angles  POA,  FOB,  is  right,  and  therefore  PO 
is  perpendicular  to  the  plane  OAB. 

650.  The  angle  between  two  intersecting  curves  is  the  angle 
between  their  tangents,  at  the  point  of  intersection. 


When  the  curves  are  arcs   of  great  circles  of  the  same 
sphere,  the  angle  is  called  a  Spherical  Angle. 


TRI-DIMENSIONAL  SPHERICS,  253 

651.  If  from  the  vertices,  A  and  F,  of  any  two  angles  in  a 
sphere,  as  poles,  great  circles,  BC  and  GH^  be  described,  the 
angles  will  be  to  one  another  in  the  ratio  of  the  arcs  of  these 
circles  intercepted  between  their  sides  (produced  if  necessary). 


For  the  angles  A  and  F  are  equal  respectively  to  the  angles 
BOC  and  GOH. 

(591.  Parallels  intersecting  the  same  plane  are  equally  inclined  to  it.) 

But  BOC  and  GOH  are  angles  at  the  centers  of  equal  cir- 
cles, and  therefore,  by  506,  are  to  one  another  in  the  ratio  of 
the  arcs  BC  and  GH. 

652.  Corollary.  Any  great-circle-arc  drawn  through  the 
pole  of  a  given  great  circle  is  perpendicular  to  that  circle. 

FG  is  ±  GH.  For,  by  hypothesis,  FG  is  a  quadrant ;  there- 
fore the  great  circle  described  with  G  as  pole  passes  through  F, 
and  so  the  arc  intercepted  on  it  between  GF  and  GH  is,  by 
648,  also  a  quadrant.  But,  by  651,  the  angle  at  G  is  to  this 
quadrant  as  a  straight  angle  is  to  a  half-line. 

Inversely,  any  great-circle-arc  perpendicular  to  a  great  cir- 
cle will  pass  through  its  pole. 

For  if  we  use  G  as  pole  when  the  angle 'at  (7  is  a  right 
angle,  then  FH,  its  corresponding  arc,  is  a  quadrant  when  GF 
is  a  quadrant ;  therefore,  by  649,  F  is  the  pole  of  GH. 


254 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  VIII. 

653.   The  smallest  line  in  a  sphere^  between  two  points,  is  the 
great-circle-arc  not  greater  than  a  semicircle^  which  joins  them. 


Hypothesis.  AB  is  a  greaf-c/rc/s-arc,  not  greater  than  a 
semicircle,  joining  any  two  points  A  and  B  on  a  sphere. 

First,  let  the  points  A  and  B  be  joined  by  the  broken  line  A  CB, 
which  consists  of  the  two  great- circle -arcs  A  C  and  CB. 
Conclusion.    AC  +  CB  >  AB. 

Proof.    Join  O,  the  center  of  the  sphere,  with  A,  By  and  C. 
^  AOC  -f  ^  COB  >  ^  AOB. 

(603.  If  three  lines  not  in  the  same  plane  meet  at  one  point,  any  two  of  the  angles 
formed  are  together  greater  than  the  third.) 

But  the  corresponding  arcs  are  in  the  same  ratio  as  these  angles, 

.-.     AC  -\-  CB>AB. 
Second,  let  F  be  any  point  whatever  on  the  great-circle-arc  AB. 
The  smallest  line  on  the  sphere  from  A  to  B  must  pass  through  B. 

For  by  revolving  the  great-circle-arcs  AB  and  BB  about  A  and  B 
as  poles,  describe  circles. 

These  circles  touch  at  B,  and  lie  wholly  without  each  other ;  for  let 
F  be  any  other  point  in  the  circle  whose  pole  is  B,  and  join  FA,  FB 
by  great-circle-arcs,  then,  by  our  First, 

FA  ^  FB>  AB, 
.*.    FA  >  FA,  and  F  lies  without  the  circle  whose  pole  is  A. 


TRI-DIMENSIONAL  SPHERICS.  255 

Now  let  ADEB  be  any  line  on  the  sphere  from  AXo  B  not  passing 
through  /*,  and  therefore  cutting  the  two  circles  in  different  points,  one 
in  D,  the  other  in  E.  A  portion  of  the  line  ADEB,  namely,  DE,  lies 
between  the  two  circles.  Hence  if  the  portion  AD  be  revolved  about 
A  until  it  takes  the  position  AGP,  and  the  portion  BE  be  revolved 
about  B  into  the  position  BHP,  the  line  AGPHB  will  be  less  than 
ADEB.  Hence  the  smallest  line  from  A  to  B  passes  through  P,  that 
is,  through  any  or  every  point  in  AB ;  consequently  it  must  be  the  arc 
AB  itself. 

654.  Corollary.  A  sect  is  the  smallest  line  in  a  plane 
between  two  points. 


BOOK    IX. 


TWO-DIMENSIONAL  SPHERICS. 

INTRODUCTION.     . 

655.  Book  IX.  will  develop  the  Geometry  of  the  Sphere, 
from  theorems  and  problems  almost  identical  with  those  whose 
assumption  gave  us  Plane  Geometry.  In  Book  VIII.,  these 
have  been  demonstrated  by  considering  the  sphere  as  contained 
in  ordinary  tri-dimensional  space.  But,  if  we  really  confine  our- 
selves to  the  sphere  itself,  they  do  not  admit  of  demonstration, 
except  by  making  some  more  difficult  assumption :  and  so  they 
are  the  most  fundamental  properties  of  this  surface  and  its 
characteristic  line,  the  great  circle ;  just  as  the  assumptions  in 
our  first  book  were  the  most  fundamental  properties  of  the 
plane  and  its  characteristic  line. 

So  now  we  will  call  a  great  circle  simply  the  spherical  line ; 
and,  whenever  in  this  book  the  word  line  is  used,  it  means 
spherical  line.  Sect  now  means  a  part  of  a  line  less  than  a 
half-line. 

257 


258  THE  ELEMENTS  OF  GEOMETRY. 


Two-Dimensional  Definition  of  the  Sphere. 

656.  Suppose  a  closed  line,  such  that  any  portion  of  it  may 
be  moved  about  through  every  portion  of  it  without  any  other 
change.  Suppose  a  portion  of  this  line  is  such,  that,  when 
moved  on  the  line  until  its  first  end  point  comes  to  the  trace  of 
its  second  end  point,  that  second  end  point  will  have  moved  to 
the  trace  of  its  first  end  point.  Call  such  a  portion  a  half- 
line,  and  any  lesser  portion  a  sect.  Suppose,  that,  while  the 
extremities  of  a  half-line  are  kept  fixed,  the  whole  line  can  be 
so  moved  that  the  slightest  motion  takes  it  completely  out  of 
its  trace,  except  in  the  two  fixed  points.  Such  motion  would 
generate  a  surface  which  we  will  call  the  Sphere. 


Fundamental  Properties  of  the  Sphere. 

ASSUMPTIONS. 

657.  A  figure  may  be  moved  about  in  a  sphere  without  any 
other  change ;  that  is,  figures  are  independent  of  their  place  on 
the  sphere. 

658.  Through  any  two  points  in  the  sphere  can  be  passed  a 
line  congruent  with  the  generating  line  of  the  sphere.  In 
Book  IX.,  the  word  line  will  always  mean  such  a  line,  and  sect 
will  mean  a  portion  of  it  less  than  half. 

659.  Two  sects  cannot  meet  twice  on  the  sphere  ;  that  is,  if 
two  sects  have  two  points  in  common,  the  two  sects  coincide 
between  those  two  points. 

660.  If  two  lines  have  a  common  sect,  they  coincide 
throughout.  Therefore  through  two  points,  not  end  points  of 
a  half-line,  only  one  distinct  line  can  pass. 

661.  A  sect  is  the  smallest  path  between  its  end  points  in 
the  sphere. 


TWO-DIMENSIONAL   SPHERICS.  259 

662.  A  piece  of  the  sphere  from  along  one  side  of  a  line 
will  fit  either  side  of  any  other  portion  of  the  line. 

DEFINITIOXS. 

663.  If  one  end  point  of  a  sect  is  kept  fixed,  the  other  end 
point  moving  in  the  sphere  describes  what  is  called  an  arc, 
while  the  sect  describes  at  the  fixed  point  what  is  called  a 
spherical  angle.  The  angle  and  arc  are  greater  as  the  amount 
of  turning  in  the  sphere  is  greater. 


664.  When  a  sect  has  turned  sufficiently  to  fall  again  into 
the  same  line,  but  on  the  other  side  of  the  fixed  point  or  ver- 
tex, the  angle  described  is  called  a  straight  angle,  and  the  arc 
described  is  called  a  semicircle. 

665.  Half  a  straight  angle  is  called  a  right  angle. 

666.  The  whole  angle  about  a  point  in  a  sphere  is  called  a 
perigon :  the  whole  arc  is  called  a  circle. 

The  point  is  called  the  pole  of  the  circle,  and  the  equal 
sects  are  called  its  spherical  radii. 

ASSUMPTIONS. 

667.  A  circle  can  be  described  from  any  pole,  with  any  sect 
as  spherical  radius. 

668.  All  straight  angles  are  equal. 


26o 


THE  ELEMENTS  OF  GEOMETRY. 


669.  Corollary  I.     All  perigons  are  equal. 

670.  Corollary  II.  If  one  extremity  of  a  sect  is  in  a  line, 
the  two  angles  on  the  same  side  of  the  line  as  the  sect  are 
together  a  straight  angle. 


671.  Corollary  III.  Defining  adjacent  angles  as  two 
angles  having  a  common  vertex,  a  common  arm,  and  not  over- 
lapping, it  follows,  that,  if  two  adjacent  angles  together  equal 
a  straight  angle,  their  two  exterior  arms  fall  into  the  same 
line. 

672.  Corollary  IV.  If  two  sects  cut  one  another,  the 
vertical  angles  are  equal. 


673.  Any  line  turning  in  the  sphere  about  one  of  its  points, 
through  a  straight  angle,  comes  to  coincidence  with  its  trace, 
and  has  described  the  sphere. 

674.  Corollary  I.     The  sphere  is  a  closed  surface. 


TWO-DIMENSIONAL  SPHERICS,  26 1 

675.  Corollary  II.     Every  line  bisects  the  sphere,  —  cuts 
it  into  hemispheres. 


Reduced  Properties  of   the  Sphere. 

Theorem   I. 

676.  All  lilies  in  a  sphere  ijttersect,  and  bisect  each  other  at 
their  points  of  intersection. 


Let  BB'  and  CC  be  any  t\vo  lines. 

Since,  by  675,  each  of  them  bisects  the  sphere,  therefore  the  second 
cannot  lie  wholly  in  one  of  the  hemispheres  made  by  the  first,  therefore 
they  intersect  at  two  points ;  let  them  intersect  at  A  and  A' . 

If  BA  C  be  revolved  in  the  sphere  about  A  until  some  point  in  the 
sect  AB  coincides  with  some  point  in  the  sect  AB' y  then  ABA'  will  lie 
along  AB'A'. 

(660.  Through  two  points,  not  end  points  of  a  half-line,  only  one  distinct  line  can 

pass.) 

Then,  also,  since  the  angles  BA  C  and  B^A  C,  being  vertical,  are, 
by  672,  equal,  AC  A'  will  lie  along  AC  A'.  Therefore  the  second  point 
of  intersection  of  AB  and  A  C  must  coincide  with  the  second  point  of 
intersection  of  AB'  and  A  C,  and  ABA'  be  a  half-line,  as  also  A  CA' . 


262 


THE  ELEMENTS   OF  GEOMETRY. 


677.  A  spherical  figure  such  as  ABA'CA^  which  is  contained 
by  two  half -lines,  is  called  a  Lune. 


Theorem   II. 

678.   The  angle  contained  by  the  sides  of  a  hme  at  one  of  their 
points  of  intersection  equals  the  angle  contained  at  the  other 


For,  if  the  two  angles  are  not  equal,  one  must  be  the  greater.  Sup- 
pose :^  A>  :^  A', 

Move  the  lune  in  the  sphere  until  point  A'  coincides  with  the  trace 
of  point  A,  and  half-line  A'BA  coincides  with  the  trace  of  half-line 
A  CA' .  Then,  since  we  have  supposed  -4-  A>  ^  A' ,  the  half-Hne  A'  CA 
would  start  from  the  trace  of  A  between  the  trace  of  AC  A'  and  the 
trace  of  ABA' .     Starting  between  them,  it  could  meet  neither  again 


TWO-DIMENSIONAL  SPHERICS. 


263 


until  it  reached  the  trace  of  A' ;  and  so  we  would  have  the  surface  of  the 

lune  less  than  its  trace,  which  is  contrary  to  our  first  assumption  (657). 

.*.     the  two  angles  cannot  be  unequal. 


DEFINITIONS. 

679.  The  Supplement  of  a  Sect  is  the  sect  bywhich  it  differs 
from  a  half-line. 


680.  One-quarter  of  a  line  is  called  a  Quadrant. 
618.  A  Spherical  Polygoii  is  a  closed  figure  in  the  sphere 
formed  by  sects. 


682.  A  Spherical  Triangle  is  a  convex  spherical  polygon  of 
three  sides. 

683.  Symmetrical  Spherical  Polygons  are  those  in  which  the 
sides  and  angles  of  the  one  are  respectively  equal  to  those  of 


264  THE  ELEMENTS  OF  GEOMETRY, 

the  other,  but  arranged  in  the  reverse  order.  If  one  end  of  a 
sect  were  pivoted  v^ithin  one  polygon,  and  one  end  of  another 
sect  pivoted  within  the  symmetrical  polygon,  and  the  two  sects 
revolved  so  as  to  pass  over  the  equal  parts  at  the  same  time, 
one  sect  would  move  clockwise,  while  the  other  moved  counter- 
clockwise. 


684.  Two  points  are  symmetrical  with  respect  to  a  fixed 
line,  called  the  axis  of  symmetry,  when  this  axis  bisects  at 
right  angles  the  sect  joining  the  two  points. 

Any  two  figures  are  symmetrical  with  respect  to  an  axis 
when  every  point  of  one  has  its  symmetrical  point  on  the 
other. 

Theorem   III. 

685.  The  perimeter  of  a  convex  spherical  polygon  wholly  con- 
tained within  a  second  spherical  polygon  is  less  than  the  perimeter 
of  the  second. 


Let  ABFGA  be  a  convex  spherical  polygon  wholly  contained  in 
ABCDEA, 


TWO-DIMENSIONAL   SPHERICS.  26$ 

Produce  the  sides  A  G  and  GF  to  meet  the  containing  perimeter  at 
K  and  L  respectively. 

By  66 1,  a  sect  is  the  smallest  path  between  its  end  points, 

.-.     BF<BCLF,     and     LG<LKG,     and     KA<KDEA; 

/.    BFGA  <  BCLGA  <  BCKA  <  BCDEA, 


Theorem   IV. 

686.    The  stint  of  the  sides  of  a  convex  spherical  polygon  is 
less  than  a  line. 


Let  ABCDEA  be  the  polygon,  and  let  its  sides  BA  and  ^C  be 
produced  to  meet  again  at  M.  Since  the  polygon  is  convex,  the 
^BCD  <?,t.  :^,  and  also  ^  BAB  <  st.  ^  ;  therefore  ABCDEA  lies 
wholly  within  ABCMA  ;  therefore,  by  685,  the  perimeter  oi  ABCDEA 
<  the  perimeter  of  ABCMA,  that  is,  less  than  a  line. 

SYMMETRY   WITHOUT   CONGRUENCE. 

687.  If  two  figures  have  central  symmetry  in  a  plane,  either 
can  be  made  to  coincide  with  the  other  by  turning  it  in  the 
plane  through  a  straight  angle.  This  holds  good  when  for 
"plane"  we  substitute  "sphere." 

If  two  figures  have  axial  symmetry  in  a  plane,  they  can  be 
made  to  coincide  by  folding  the  plane  over  along  the  axis,  but 
not  by  any  sliding  in  their  plane.     That  is,  we  must  use  the 


266  THE  ELEMENTS   OF  GEOMETRY. 

third  dimension  of  space,  and  then  their  congruence  depends 
on  the  property  of  the  plane  that  its  two  sides  are  indistin- 
guishable, so  that  any  piece  will  fit  its  trace  after  being  turned 
over.  This  procedure,  folding  along  a  line,  can  have  no  place 
in  a  strictly  two-dimensional  geometry ;  and,  were  we  in  tri- 
dimensional spherics,  we  could  say,  that  from  the  outside  a 
sphere  is  convex,  while  from  the  inside  it  is  concave,  and  that 
a  piece  of  it,  after  being  turned  over,  will  not  fit  its  trace,  but 
only  touch  it  at  one  point.  So  figures  with  axial  symmetry, 
according  to  684,  on  a  sphere  cannot  be  made  to  coincide ;  and 
the  word  symmetrical  is  henceforth  devoted  entirely  to  such. 


Theorem  V. 

688.  Two  spherical  tria7tgles  having  two  sides  and  the  in- 
cluded angle  of  one  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other,  are  either  congruent  or  symmetrical. 


(i)  If  the  parts  given  equal  are  arranged  in  the  same  order,  as  in 
DEF  and  ABC,  then  the  triangle  DEF  can  be  moved  in  the  sphere 
until  it  coincides  with  ABC. 

(2)  If  the  parts  given  equal  are  arranged  in  reverse  order,  as  in 
DEF  and  A' B'  C\  by  making  a  triangle  symmetrical  to  A'B'C,  as 
ABC,  we  get  the  equal  parts  arranged  in  the  same  order  as  in  DEF, 
which  proves  DEF  congruent  to  any  triangle  symmetrical  to  A'B'  C\ 


TWO-DIMENSIONAL   SPHERICS. 


267 


Theorem  VI. 

689.  If  two  spherical  triangles  have  two  sides  of  the  one 
equal  respectively  to  two  sides  of  the  other,  but  the  ijicluded  angle 
of  tJie  first  greater  than  the  included  angle  of  the  second,  then 
the  third  side  of  the  first  will  be  greater  than  the  third  side  of 
the  second. 


Hypothesis.    AB  =  DE,  and  AC  ^  EF,  bui  4.  BAC>  4.  EDF, 

Conclusion.    BC  >  EF. 

Proof.  When  AB  coincides  with  DE,  if  the  points  C  and  F  are 
on  the  same  side  of  the  line  AB,  then,  since  ^  BAC  >  ^  EDF,  the 
side  DF  will  lie  between  BA  and  A  C,  and  DF  must  stop  either  within 
the  triangle  ABC,  on  BC,  or  after  cutting  BC. 

When  F  is  within  the  triangle,  by  685,  ^C  +  CA  >  EF  +  FD ; 
but^C=  DF, 

.-.    BC>EF, 

When  F  lies  on  BC,  then  EF  is  but  a  part  of  BC.  In  case  DF 
cuts  BC,  call  their  intersection  point  G.  Then  BG  +  GF  >  EF,  and 
GC  -^  GD>  AC, 

.-.    BC  ^  FD  >AC  ^  EF', 
hwi  FD  =  AC, 

.'.    BC>EF. 

If,  when  one  pair  of  equal  sides  coincide,  the  other  pair  lie  on  oppo- 
site sides  of  the  line  of  coincidence,  the  above  proof  will  show  the  third 
side  of  a  triangle  symmetrical  to  the  first  to  be  greater  than  the  third 
side  of  the  second  triangle,  and  therefore  the  third  side  of  the  first 
greater  than  the  third  side  of  the  second. 


268  THE  ELEMENTS  OF  GEOMETRY. 

690.  From  6'^%  and  689,  by  33,  Rule  of  Inversion,  if  two 
spherical  triangles  have  two  sides  of  the  one  equal  respectively 
to  two  sides  of  the  other,  the  included  angle  of  the  first  is 
greater  than,  equal  to,  or  less  than,  the  included  angle  of  the 
second,  according  as  the  third  side  of  the  first  is  greater  than, 
equal  to,  or  less  than,  the  third  side  of  the  second. 

691.  Corollary.  Therefore,  by  688,  two  spherical  triangles 
having  three  sides  of  the  one  equal  respectively  to  three  sides 
of  the  other  are  either  congruent  or  symmetrical. 


Problem   I. 
692.   To  bisect  a  given  spherical  angle. 


Let  BA  C  be  the  given  2^ . 

In  its  arms  take  equal  sects  AB  and  A  C  each  less  than  a  quadrant. 

Join  BC.  With  B  as  pole,  and  any  sect  greater  than  half  BC,  but 
not  greater  than  a  quadrant,  as  a  spherical  radius,  describe  the  circle 
EDF.  With  an  equal  spherical  radius  from  C  as  pole  describe  an  arc 
intersecting  the  circle  EDF  at  D  within  the  lune  BAC.  Join  DB, 
DC,  DA.     i9^  shall  bisect  the  angle  ^^C. 

For  the  'as  ABD,  A  CD,  having  AD  common  and  AB  =  AC,  and 
BD  =  DC  (spherical  radii  of  equal  circles),  are,  by  691,  symmetrical, 

.'.     ^  BAD  =  ^  CAD. 


TWO-DIMENSIONAL  SPHERICS. 


269 


693.  Corollary  I.  To  bisect  the  re-entrant  angle  BAC, 
produce  the  bisector  of  the  angle  BAC. 

694.  Corollary  II.  To  erect  a  perpendicular  to  a  given 
line  from  a  given  point  in  the  line,  bisect  the  straight  angle  at 
that  point. 


Problem  II. 
695.   To  bisect  a  given  sect. 


With  A  and  B,  the  extremities  of  the  given  sect,  as  poles,  and  equal 
spherical  radii  greater  than  half  AB,  but  less  than  a  quadrant,  describe 
arcs  intersecting  at  C. 

Join  AC  and  BC;  and,  by  692,  bisect  the  angle  ACB,  and  produce 
the  bisector  to  meet  AB  at  Z>. 

B>  is  the  mid  point  of  the  sect  AB. 

For,  by  688,  a"  A  CD  is  symmetrical  to  'aBCJD, 

696.  Corollary.  The  angles  at  the  base  of  an  isosceles 
spherical  triangle  are  equal. 


270 


THE  ELEMENTS  OF  GEOMETRY. 


Problem   III. 

697.   To  draw  a  perpendicular  to  a  given  line  from  a  given 
point  in  the  sphere  not  in  the  line. 


Given,  the  line  AB,  and  point  C. 

Take  a  point  D  on  the  other  side  of  the  line  AB,  and  with  C  as 
pole,  and  CD  as  spherical  radius,  describe  an  arc  cutting  AB  in  F 
and  G.     Bisect  the  sect  FG  at  B,  and  join  CB.     CB  shall  be  ±  AB. 

For,  by  691,  'as  GCB  and  FCB  are  symmetrical. 


Theorem  VII. 

698.  If  two  lines  be  drawn  in  a  sphere,  at  right  angles  to  a 
given  line,  they  will  intersect  at  a  point  from  which  all  sects 
drawn  to  the  given  line  are  equal. 


Let  AB  and  CB,  drawn  at  right  angles  to  AC,  intersect  at  B,  and 
meet  A  C  again  at  A'  and  C  respectively. 


TWO-DIMENSIONAL  SPHERICS.  27 1 

Then  4.  BA'C  =  ^  BAC,       and       4.  EC  A'  =  4  BCA' -, 

(678.  The  angles  contained  by  the  sides  of  a  lune,  at  their  two  points  of  intersection, 

are  equal.) 
moreover, 

AC  =  A'C\ 

for  they  have  the  common  supplement  A  C .  Hence,  keeping  A  and  C 
on  the  line  AC,  slide  ABC  until  AC  comes  into  coincidence  with  A' C . 
Then,  the  angles  at  A,  C,  A\  C,  being  all  right,  AB  will  lie  along  A'B, 
and  CB  along  C'B,  and  hence  the  figures  AB  C  and  A'B  C  coincide. 

.-.     each  of  the  half-Unes  ABA'  and  CB  C  is  bisected  at  B. 

In  like  manner,  any  other  line  drawn  at  right  angles  to  ^  C  passes 
through  B,  the  mid  point  of  ABA' . 

Hence  every  sect  from  AC  to  B  is  a.  quadrant. 

699.  Corollary   I.     A   line   is   a   circle   whose   spherical 
radius  is  a  quadrant. 

700.  Corollary   II.     A  point  which  is  a  quadrant  from 
two  points  in  a  line,  and  not  in  the  line,  is  its  pole. 

701.  Corollary  III.     Any  sect  from  the  pole  of  a  line  to 
the  line  is  perpendicular  to  it. 

702.  Corollary  IV.     Equal  angles  at  the  poles  of  lines 
intercept  equal  sects  on  those  lines. 

703.  Corollary  V.     If  K  be  the  pole,  and  FG  a  sect  of 


any  other  line,  the  angles  and  semilunes  ABC  and  FKG  are  to 
one  another  as  ^  6^  to  FG. 


2/2 


THE  ELEMENTS  OF  GEOMETRY. 


704.  We  see,  from  698,  that  a  spherical  triangle  may  have 
two  or  even  three  right  angles. 

If  a  spherical  triangle  ABC  has  two  right  angles,  B  and  C^ 
it  is  called  a  bi-rectaiigular  triangle.  By  698,  the  vertex  A  is 
the  pole  of  BC,  and  therefore  AB  and  AC  are  quadrants. 

705.  The  Polar  of  a  given  spherical  triangle  is  a  spherical 
triangle,  the  poles  of  whose  sides  are  respectively  the  vertices 
of  the  given  triangle,  and  its  vertices  each  on  the  same  side  of 
a  side  of  the  given  triangle  as  a  given  vertex. 


Theorem  VIII. 

706.    If,  of  two  spherical  triangles,  the  first  is  the  polar  of 
the  second,  tJien  the  second  is  the  polar  of  the  first. 


Hypothesis.    Lei  A'B'C  be  the  polar  of  ABC, 
Conclusion.    Then  ABC  is  the  polar  of  A'B'C. 


TWO-DIMENSIONAL  SPHERICS,  2/3 

Proof.    Join  A'B  and  A' C, 

Since  B  is  the  pole  of  A'  C ,  therefore  BA'  is  a  quadrant ;  and  since 
C  is  the  pole  of  A'B\  therefore  CA'  is  a  quadrant ; 

.*.    by  700,  A'  is  the  pole  of  BC. 

In  like  manner,  B'  is  the  pole  of  A  C,  and  C  of  AB, 
Moreover,  A  and  A'  are  on  the  same  side  oi  B'C'y  B  and  B'  on  the 
same  side  oi  A'C,  and  C  and  C  on  the  same  side  oi  A'B\ 

.-.    ^.5C  is  the  polar  of  A'B'C, 


Theorem  IX. 

707.  In  a  pair  of  polar  triangles,  any  angle  of  either  inter- 
ceptSy  on  the  side  of  the  other  which  lies  opposite  to  it,  a  sect  which 
is  the  supplement  of  that  side. 


Let  ABC  and  A'B^C  be  two  polar  triangles. 

Produce  A'B'  and  B' C  to  meet  BC  at  Z>  and  ^  respectively. 
Since  B  is  the  pole  of  B'  C ,  therefore  BE  is  a  quadrant ;  and  since  C 
is  the  pole  of  A'B' ,  therefore  CD  is  a  quadrant ;  therefore  BE  -f  CD 
=  half-line,  but  BE  -{-  CD  =  BC  -{-  DE.  Therefore  DE,  the  sect 
of  BC  which  A'  intercepts,  is  the  supplement  of  BC. 

Exercises.  109.  Any  lune  is  to  a  tri-rectangular  triangle 
as  its  angle  is  to  half  a  right  angle. 


274 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  X. 

708.  If  two  angles  of  a  spherical  triangle  be  equals  the  sides 
which  subtend  them  are  equal. 


Hypothesis.    In  ^  ABC  let  4.  A  =  ^  C, 

Conclusion.    BC  =  AB. 
Proof.     For  draw  A'B'C,  the  polar  of  ABC. 
Now,  on  B' C  and  A'B'  the  equal  7^^  A  and  C  intercept  equal  sects. 
Therefore  B'  C  and  A'B'y  being,  by  707,  the  supplements  of  these 
equal  sects,  are  equal,  .*.     "4-  A'  —  -^  C, 

(696.  The  angles  at  the  base  of  an  isosceles  spherical  triangle  are  equal.) 

.'.     the  supplements  oi  BC  and  AB  are  equal, 
/.      BC  ^  AB, 


Theorem   XI. 

709.  If  one  angle  of  a  spherical  triangle  be  greater  than  a 
second^  the  side  opposite  the  first  must  be  greater  than  the  side 
opposite  the  second. 


In'^ABC,  4  C>4.A.        Make  ^  ACD  =  ^^; 
/.    by  708,    AD  =  CD, 


TWO-DIMENSIONAL  SPHERICS.  2y$ 

But,  by  66i,  CD  -^  DB>  BC, 

.-.    AD  -^  DB>BC. 

710.  From  708  and  709,  by  33,  Rule  of  Inversion, 

If  one  side  of  a  spherical  triangle  be  greater  than  a  second, 
the  angle  opposite  the  first  must  be  greater  than  the  angle 
opposite  the  second. 

Theorem   XII. 

711.  Two  spherical  triangles  having  two  angles  and  the 
included  side  of  the  one  equal  respectively  to  two  angles  and 
the  included  side  of  the  other^  are  either  congrue7it  or  symmetricaL 


For  the  first  triangle  can  be  moved  in  the  sphere  into  coincidence 
with  the  second,  or  with  a  triangle  made  symmetrical  to  the  second. 


Theorem   XIII. 

712.  Two  spherical  triangles  having  three  angles  of  the  one 
equal  respectively  to  three  angles  of  the  other^  are  either  congruetit 
or  symmetrical. 

Since  the  given  triangles  are  respectively  equiangular,  their  polars 
are  respectively  equilateral. 

(702.  Equal  angles  at  the  poles  of  lines  intercept  equal  sects  on  those  lines;  and, 
by  707,  these  equal  sects  are  the  supplements  of  corresponding  sides.) 


2/6  THE  ELEMENTS  OF  GEOMETRY. 

Hence  these  polars,  being,  by  691,  congruent  or  symmetrical,  are 
respectively  equiangular,  and  therefore  the  original  spherical  triangles 
are  respectively  equilateral. 

Theorem  XIV. 

713.  An  exterior  angle  of  a  spherical  triangle  is  greater  than^ 
eqnal  to,  or  less  tkaUy  either  of  the  interior  opposite  angles,  accord- 
ing as  the  medial  from  the  other  interior  opposite  angle  is  less 
than,  equal  to,  or  greater  than,  a  quadrant. 


Let  ACD  be  an  exterior  angle  of  the  a"  ABC.     By  695,  bisect  AC 
at  E.     Join  BE,  and  produce  to  E,  making  EE  =  BE.    Join  EC» 

A  ABE  ^  A  CEE, 
(688.  Spherical  triangles  having  two  sides  and  the  included  angle  equal  are  congru- 
ent or  symmetrical.) 

.'.     ^  BAE  =  ^  ECE. 
If,  now,  the  medial  BE  be  a  quadrant,  BEE  is  a  half-line,  and,  by 
676,  E  lies  onBD; 

.'.     4.DCE  coincides  with  ^  ECE, 
.\     ^  DCE  =  t  BAE. 
If  the  medial  BE  be  less  than  a  quadrant,  BEE  is  less  than  a  half- 
line,  and  E  lies  between  A  C  and  CD ; 

.♦.     ^  DCA  >  ^  ECE, 
.-.     ^  DCA  >  ^  BAC. 
And  if  BE  be  greater  than  a  quadrant,  BEE  is  greater  than  a  half- 
hne,  and  E  lies  between  CD  and  A  C  produced ; 
.-.     ^  DCA  <   -4.  ECE, 
/.     ^  DCA   <  4  BAC. 


TWO-DIMENSIONAL   SPHERICS.  2'J'J 

Thus,  according  as  BE  is  greater  than,  equal  to,  or  less  than,  a 
quadrant,  the  exterior  ^  A  CD  is  less  than,  equal  to,  or  greater  than,  the 
interior  opposite  2(1  BA  C, 

714.  From  713,  by  33,  Rule  of  Inversion, 

According  as  the  exterior  angle  ACD  is  greater  than,  equal 
to,  or  less  than,  the  interior  opposite  angle  BAC,  the  medial 
B£  is  less  than,  equal  to,  or  greater  than,  a  quadrant. 


Theorem   XV. 

715.  Two  spherical  triangles  having  the  a^tgles  at  the  base  of 
the  07ie  eqical  to  the  a^tgles  at  the  base  of  the  other^  and  the  sides 
opposite  one  pair  of  equal  angles  eqical,  are  either  coJigricent  or 
symmetrical,  provided  that  in  neither  triangle  is  the  third 
angular  point  a  quadrant  from  any  point  iii  that  half  of  its 
base  not  adjaceftt  to  07ie  of  the  sides  equal  by  hypothesis. 


First,  if  the  parts  given  equal  lie  clockwise  in  the  two  spherical 
triangles,  as  in  ABC  and  DEF,  where  we  suppose  ^  B  =  }^  E, 
4.  C  =  4.  F,2ir\dAB  ==  DE,  make  DE  coincide  with  AB ;  then  EF 
will  lie  along  BC,  and  DF  must  coincide  with  AC.  For  if  it  could 
take  any  other  position,  as  AG,  it  would  make  2^^  AGC  with  exterior 
^  AGB  =  interior  opposite  ^  C,  and  therefore,  by  714,  with  medial 
AH  a  quadrant,  which  is  contrary  to  our  hypothesis. 

Second,  if  the  equal  parts  lie  in  one  spherical  triangle  clockwise, 
in  the  other  counter-clockwise,  as  in  'KA'B'C  and  ^  DEF,  then 
A  B>EF  ^^  ABC,  which  is  symmetrical  to  a  A' B' C, 


2/8  THE  ELEMENTS  OF  GEOMETRY. 


Theorem   XVI. 

716.  If  a  sect  drawn  in  the  sphere  from  a  point  perpendictilar 
to  a  line  be  less  than  a  qnadranty  it  is  the  smallest  which  can  be 
drawn  in  the  sphere  from  the  point  to  the  line ;  of  others^  that 
which  is  nearer  the  perpendicular  is  less  than  that  which  is  more 
remote  ;  also  to  every  sect  drawn  on  one  side  of  the  perpendicular 
there  can  be  drawn  one^  and  only  one^  sect  equal  on  the  other  side. 


Let  A  be  the  point,  BD  the  Hne,  AB  J.  BD^  and  AD  nearer  than 
AE  to  AB. 

Produce  AB  to  C,  making  BC  ^  AB.    Join  CD,  CE, 

Then  AD  =  CD. 

(688.  Spherical  triangles  having  two  sides  and  the  included  angle  equal  are  con- 
gruent or  symmetrical.) 

But,  by  661, 

AD  -j-  CD  or  iAD  >  AC  or  2AB, 

.-.     AD  >  AB. 
Also,  by  685, 

AE  +  CE  or  2AE  >  AD  -f  CD  or  2 AD. 

Again,  make  BE  —  BD.    Join  AE. 

As  before,  AE  =  AD,  and  we  have  already  shown  that  no   two 
sects  on  the  same  side  of  the  perpendicular  can  be  equal. 


TWO-DIMENSIONAL   SPHERICS.  279 

717.  Corollary.    The  greatest  sect  that  can  be  drawn  from 
A  to  BD  is  the  supplement  of  AB. 


Theorem   XVII. 

718.  If  two  spherical  triangles  have  two  sides  of  the  one 
equal  to  two  sides  of  the  other^  and  the  angles  opposite  one  pair 
of  equal  sides  equals  the  angles  opposite  the  other  pair  a7'e  either 
equal  or  supplemental. 


First,  given  the  equal  parts  AB  =  DE,  AC  =  DF,  and  ^  B  = 
^  E  arranged  clockwise  in  the  two  spherical  triangles. 

\i  -2^  A  =.  }^  D,  the  spherical  triangles  are  congruent. 

\i  ^  A  is  not  =  ^  Z>,  one  must  be  the  greater. 

Suppose  -4.  A>  -i^  D.  Make  DE  coincide  with  AB.  Then,  since 
^B  =  :4E,  side  EF  will  he  along  BC;  since  ^  A>  ^  D,  side  VF 
will  lie  between  AB  and  A  C,  as  at  A  G. 

Now,  the  two  angles  at  G  are  supplemental ;  but  one  is  -^  F  and 
the  other  =  ^  C,  because  "a  CFG  is  isosceles. 

Second,  if  the  parts  given  equal  lie  clockwise  in  one  spherical  tri- 
angle and  counter-clockwise  in  the  other,  as  in  "a  A'B'  C  and  'a  DEF, 
then,  taking  the  "a  ABC  symmetrical  to  A'B' C,  the  above  proof  shows 
^  F  equal  or  supplemental  to  ^  C,  that  is,  to  ^  C\ 


280  THE  ELEMENTS  OF  GEOMETRY. 

Theorem   XVIII. 
719.  Symmetrical  isosceles  spherical  triangles  are  congruent. 


For,  since  four  sides   and   four  angles   are   equal,  the  distinction 
between  clockwise  and  counter-clockwise  is  obliterated. 


Theorem   XIX. 

720.  The  locus  of  a  point  from  which  the  two  sects  drawn  to 
two  given  points  are  equal  is  the  line  bisecting  at  right  angles 
the  sect  joining  the  two  given  points. 


For  every  point  in  this  perpendicular  bisector,  and  no  point  out  of 
it,  possesses  the  property. 


TWO-DIMENSIONAL  SPHERICS. 


281 


Problem  IV. 

721.   To  pass  a  circle  through  auy  three  points ,  or  to  find  the 
circumcenter  of  any  spherical  triangle. 


Find  the  intersection  point  of  perpendiculars  erected  at  the  mid- 
points of  two  sides. 


Theorem  XX. 

722.  Any  angle  made  with  a  side  of  a  spherical  triangle  by 
joining  its  extremity  to  the  circwnc enter ^  equals  half  the  angle- 
sum  less  the  opposite  aiigle  of  the  triangle. 


FoT^A-}-^B-{-^C=2^  OCA  +  2  ^  OCB  ±  2  ^  OAB, 


^A  4-  ^B  4-  ^  C 
t  OCA  =  ^ \ U  OCB  ±  ^  OAB) 

^A  -\-  ^B  +  4.C 


^  ^B. 


282  THE  ELEMENTS  OF  GEOMETRY. 

723.  Corollary.  Symmetrical  spherical  triangles  are  equiv- 
alent. 

For  the  three  pairs  of  isosceles  triangles  formed  by  joining 
the  vertices  to  the  circumcenters  having  respectively  a  side  and 
two  adjacent  angles  equal,  are  congruent. 


Theorem  XXI. 

724.  When  three  lines  mutually  intersecty  the  two  triangles  on 
opposite  sides  of  any  vertex  are  together  equivalent  to  the  lune 
with  that  vertical  angle. 


"aABC-^-^  ADF  =  lune  ABHCA. 

For  DF  —  BC,  having  the  common  supplement  CD )  and  FA  — 
CH,  having  the  common  supplement  HF'^  and  AD  =  BH^  having  the 
common  supplement  HD ; 

.-.     aADF^aBCFT, 
.-.     A  ABC  +  A  ADF  =  A  ABC  -f  a  BCH  =  lune  ABHCA. 

725.  The  Spherical  Excess  of  a  spherical  triangle  is  the 
excess  of  the  sum  of  its  angles  over  a  straight  angle.  In  gen- 
eral, the  spherical  excess  of  a  spherical  polygon  is  the  excess 
of  the  sum  of  its  angles  over  as  many  straight  angles  as  it  has 
sides,  less  two. 


TWO-DIMENSIONAL  SPHERICS.  283 


Theorem  XXII. 

726.  A  spherical  triangle  is  equivalent  to  a  lime  whose  angle 
is  half  the  t7  iattgles  spherical  excess. 


Proof.  Produce  the  sides  of  the  "a  ABC  until  they  meet  again, 
two  and  two,  at  Dy  F^  and  H.  The  a"  ABC  now  forms  part  of  three 
lunes,  whose  angles  are  A,  Bj  and  C  respectively. 

But,  by  724,  lune  with  2^  ^  =  a  ABC  +  a  ADF. 

Therefore  the  lunes  whose  angles  are  Ay  By  and  C,  are  together 
equal  to  a  hemisphere  plus  twice  'a  ABC. 

But  a  hemisphere  is  a  lune  whose  angle  is  a  straight  angle, 

.-.     2^ ABC  =  lune  whose  ^  is  (^  +  ^  +  C  -  st.  ^) 

=  lune  whose  ^  is  e, 

727.  Corollary  I.  The  sum  of  the  angles  of  a  spherical 
triangle  is  greater  than  a  straight  angle  and  less  than  3  straight 
angles. 

728.  Corollary  II.  Every  angle  of  a  spherical  triangle  is 
greater  than  ^e. 

T2.g,  Corollary  III.  A  spherical  polygon  is  equivalent  to 
a  lune  whose  angle  is  half  the  polygon's  spherical  excess. 

730.  Corollary  IV.  Spherical  polygons  are  to  each  other 
as  their  spherical  excesses,  since,  by  703,  lunes  are  as  their 
angles. 


284  THE  ELEMENTS  OF  GEOMETRY. 

731.  Corollary  V.  To  construct  a  lune  equivalent  to  any 
spherical  polygon,  add  its  angles,  subtract  {71  —  2)  straight 
angles,  halve  the  remainder,  and  produce  the  arms  of  a  half 
until  they  meet  again. 

Theorem   XXIII. 

732.  If  the  line  be  completed  of  which  the  base  of  a  given 
spherical  triangle  is  a  sect^  and  the  other  two  sides  of  the  triangle 
be  produced  to  meet  this  line,  and  a  circle  be  passed  through  these 
tzvo  points  of  intersectio7i  and  the  vertex  of  the  triangle,  the  locns 
of  the  vertices  of  all  triangles  equivalent  to  the  give?z  triajtgle^ 
and  on  the  same  base  with  it,  and  on  the  same  side  of  that  base, 
is  the  arc  of  this  circle  terminated  by  the  intersection  points  and 
cojttaining  the  vertex. 

C 


Let  ABC  be  the  given  spherical  triangle,  AC  its  base.  Produce 
AB  and  CB  to  meet  AC  produced  in  D  and  E  respectively.  By  719, 
pass  a  circle  through  B,D,E.  Let  F  be  any  point  in  the  arc  EBD. 
Join  AP  and  CP.  AP  produced  passes  through  the  opposite  point 
D,  and  CP  through  E,  forming  with  DE  the  a  PDE.  Join  F,  its 
circumcenter,  with  P,  D,  and  E,  Since,  by  678,  the  two  angles  of  a 
lune  are  equal, 

.*.     2^:  PAC  =  St.  ^  -  :^  PDE, 

^  PCA  =  St.  ^  -  2<  PED, 
4.APC  ^  4.DPE', 


TWO-DIMENSIONAL  SPHERICS.  285 

/.    4.{PAC  ■\-  PCA  4-  APC)  =  2St.  ^s  -  ^  {PDE  +  PED  -  DPE) 

=  2  St.  ^  s  --  2  ^  EZ>E  =  a  constant. 

(722.  Any  angle  made  with  a  side  of  a  spherical  triangle  by  joining  its  extremity 
to  the  circumcenter,  equals  half  the  angle-sum  less  the  opposite  angle  of  the 
spherical  triangle.) 

733.  In  a  sphere  a  line  is  said  to  touch  a  circle  when  it 
meets  the  circle,  but  will  not  cut  it. 


Theorem  XXIV. 

734.   T/ie  line  drawn  at  right  angles  to  the  spherical  radius  of 
a  circle  at  its  extremity  touches  the  circle. 


Let  BD  be  perpendicular  to  the  spherical  radius  AB,    Join  A  with 
any  point  C  in  BD. 

By  7 1 6,  ^  C  >  AB,  therefore  C  is  without  the  circle. 
And  no  other  line  through  B,  as  BF,  can  be  tangent. 
For  draw  AE  _L  BF.     By  716,  AB  >  AE, 

,'.    E  is  within  the  circle. 


286 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  XXV. 

735.  In  a  sphere  the  stun  of  one  pair  of  opposite  angles  of  a 
quadfilateral  inscribed  in  a  circle  equals  the  sum  of  the  other 
pair. 


Join  E^  the  circumcenter,  with  A,  By  C,  D,  the  vertices  of  the 
inscribed  quadrilateral. 

By  696,  ^  ABC  =  4.  BAE  +  ^  BCE,  and  4.  ADC  =  4.  DAE 
+  4  DCE, 

/.     4.  ABC  +  4  ADC  =  4.  BAD  +  4  BCD, 


Theorem   XXVI. 
736.  In  equal  circles^  equal  angles  at  the  pole  stand  on  equal 


arcs. 


For  the  figures  may  be  slidden  into  coincidence. 


p 


TWO-DIMENSIONAL  SPHERICS. 


287 


Theorem   XXVII. 

737.   EquQ,l  spherical  chords  cut  equal  circles  into  the  same 
two  arcs. 


Theorem   XXVIII. 

738.  In  equal  circles^  angles  at  the  corresponding  poles  have 
the  same  ratio  as  their  arcs. 


Theorem  XXIX. 

739'  Four  pairs  of  equal  circles  ca7t  be  drawn  to  touch  three 
non-concurrent  lines  in  a  sphere. 


BOOK   X. 


POLYHEDRONS. 
740.  A  Polyhedron  is  a  solid  bounded  by  planes. 


741.  The  bounding  planes,  by  their  intersections,  determine 
the  Faces  of  the  polyhedron,  which  are  polygons. 

742.  The  Edges  of  a  polyhedron  are  the  sects  in  which  its 
faces  meet. 


290  THE  ELEMENTS  OF  GEOMETRY. 

743.  The  Summits  of  a  polyhedron  are  the  points  in  which 
its  edges  meet. 

744.  A  Plane  Section  of   a  polyhedron  is  the   polygon  in 
which  a  plane  passing  through  it  cuts  its  faces. 

745.  A  Pyramid  is  a  polyhedron  of  which  all  the  faces, 
except  one,  meet  in  a  point. 


746.  The  point  of  meeting  is  called  the  Apex^  and  the  face 
not  passing  through  the  apex  is  taken  as  the  Base. 

747.  The  faces  and  edges  which  meet  at  the  apex  are  called 
Lateral  Faces  and  Edges. 

748.  Two  polygons  are  said  to  be  parallel  when  each  side  of 
the  one  is  parallel  to  a  corresponding  side  of  the  other. 

749.  A  Prism  is  a  polyhedron  two  of  whose  faces  are  con- 
gruent parallel  polygons,  A»d  the  other  faces  are  parallelo- 
grams. 

750.  The  Bases  of  a  prism  are  the  congruent  parallel  poly- 
gons. 

751.  The  Lateral  Faces  of  a  prism  are  all  except  its  bases. 

752.  The  Lateral  Edges  are  the  intersections  of  the  lateral 
faces. 

753.  A  Right  Section  of  a  prism  is  a  section  by  a  plane  per- 
pendicular to  its  lateral  edges. 


POL  YHEDRONS. 


291 


754.  The  Altitude  of  a  Prism  is  any  sect  perpendicular  to 
both  bases. 


755.  The  Altitude  of  a  Pyramid  is  the  perpendicular  from 
its  vertex  to  the  plane  of  its  base. 

756.  A  Right  Prism  is  one  whose  lateral  edges  are  perpen- 
dicular to  its  bases. 


\ 


757.  Prisms  not  right  are  oblique. 

758.  A  Parallelopiped  is  a  prism  whose  bases  are  parallelo- 
grams. 

759.  A  Quader  is  a  parallelopiped  whose  six  faces  are  rect- 


angles. 


760.  A  Citbe  is  a  quader  whose  six  faces  are  squares. 


292 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   I. 

761.  All  the  summits  of  any  polyhedron  may  be  joined  by 
one  closed  line  breaking  only  in  themy  and  lying  wholly  on  the 
surface. 


For,  starting  from  one  face,  ABC .  .  .  ,  each  side  belongs  also  to  a 
neighboring  polygon. 

Therefore,  to  join  A  and  B,  we  may  omit  AB,  and  use  the  remain- 
der of  the  perimeter  of  the  neighboring  polygon  a.  In  the  same  way, 
to  join  B  and  C,  we  may  omit  BC,  and  use  the  remainder  of  the 
perimeter  of  the  neighboring  polygon  b,  unless  the  polygons  a  and  b 
have  in  common  an  edge  from  B.  In  such  a  case,  draw  from  ^  in  ^ 
the  diagonal  nearest  the  edge  common  to  a  and  b  ;  take  this  diagonal 
and  the  perimeter  of  b  beyond  it  around  to  C,  as  continuing  the  broken 
hne ;  and  proceed  in  the  same  way  from  C  around  the  neighboring 
polygon  c. 

When  this  procedure  has  taken  in  all  summits  in  faces  having  an 
edge  in  common  with  ABC  .  .  .  ,  we  may,  by  proceeding  from  the 
closed  broken  line  so  obtained,  in  the  same  way  take  in  the  summits 
on  the  next  series  of  contiguous  faces,  etc. 

So  continue  until  the  single  closed  broken  Hne  goes  once,  and  only 
once,  through  every  summit. 


POL  YHEDRONS. 


593 


Theorem   II. 

762.  Cutting  by  diagonals  the  faces  not  triangles  into  tri- 
angleSy  the  whole  surface  of  aiiy  polyhedron  contains  four  less 
triattgles  than  double  its  finmber  of  summits. 


For,  joining  all  the  summits  by  a  single  closed  broken  line,  this  cuts 
the  surface  into  two  bent  polygons,  each  of  which  contains  S  —  2  tri- 
angles, where  S  is  the  number  of  summits. 

763.  Corollary.  The  sum  of  all  the  angles  in  the  faces 
of  any  polyhedron  is  as  many  perigons  as  the  polyhedron  has 
summits,  less  two. 

764.  Remark.  Theorem  II.  is  called  Descartes'  Theorem, 
and  is  really  the  fundamental  theorem  on  polyhedrons,  though 
this  place  has  long  been  held  by  Theorem  III.,  called  Euler's 
Theorem,  which  follows  from  it  with  remarkable  elegance. 


294 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   III. 

765.    The  number  of  faces  and  summits  iri  any  polyhedron^ 
taken  together,  exceeds  by  two  the  member  of  its  edges. 


Case  I.     If  all  the  faces  are  triangles.    Then,  by  762, 

7^  =  2(6'-  2). 
But  also 

2E  =  zF, 

for  each  edge  belongs  to  two  faces,  and  so  we  get  a  triangle  for  every 
time  3  is  contained  in  2E. 

By  adding,  we  have  2E  =  2F  +  2{S  —  2);  that  is, 

F  -\-  S  =  E  -^  2, 
Case  II.     If  not  all  the  faces  are  triangles. 


Since  to  any  pyramidal  summit  go  as  many  faces  as  edges,  we  may 
replace  any  polygonal  face  by  a  pyramidal  summit  without  changing  the 


POL  YHEDRONS. 


295 


equality  or  inequality  relation  oi  F  -\-  S  \.o  E  -{■  2  ;  for  such  replace- 
ment only  adds  the  same  number  to  F  as  to  E^  and  changes  one  face 
to  a  summit.  But,  after  all  polygonal  faces  have  been  so  replaced, 
F  ■\-  S  =■  E  -\-  2y  \yy  Case  I.  Therefore  always  the  relation  was 
equality. 


Theorem  IV. 

766.    Qtiaders  having  Congruent  bases  are  to  each  other  as 
their  altitudes. 


1 

jm~-  -  -  — 

*'  \ 
1 

y 

7 

at 

A           / 

A 1 

-•r 

hH 

,>-.- 

7 

Hypothesis.  Let  a  and  c^  be  the  aliiiudes  of  two  quaders,  Q 
and  Q,  having  congruent  bases  B. 

Conclusion.     Q  :  Q  w  a  \  a'. 

Proof.  Of  a  take  any  multiple,  ma ;  then  the  quader  on  base  B 
with  altitude  ma  is  mQ. 

In  the  same  way,  take  equimultiples  na' ,  nQ. 

According  as  m  Q  is  greater  than,  equal  to,  or  less  than,  n  Q,  we 
have  ma  greater  than,  equal  to,  or  less  than  na'  \  therefore,  by  defini- 
tion, 

Q  \   Q  \'.  a  \  a\ 


Exercises,  no.  In  no  polyhedron  can  triangles  and  three- 
faced  summits  both  be  absent ;  together  are  present  at  least 
eight.  Not  all  the  faces  nor  all  the  summits  have  more  than 
five  sides. 

III.  There  is  no  seven-edged  polyhedron. 


296 


THE  ELEMENTS   OF  GEOMETRY. 


Theorem  V. 

767.  Quaders  having  equal  altitudes  are  to  each  other  as  their 
bases. 


"/HA  A 


A           A 

a 

1 
I 
1 
1 

/ 
f 

Jo- 

JP 

V 

A        / 

i 

Hypothesis.    Let  the  rectangles  he  and  h'<f  be  the  bases  of  two 
quaders,  Q  and  Q,  of  altitude  a. 

Conclusion.     Q  \  Q  w  be  \  b'(f , 

Proof.     Make  P  a  third  quader  of  altitude  a  and  base  hc\ 

Now,  considering  the  rectangles  ab  as  the  bases  of  Q  and  P,  by  766, 

Q  ',  P  '.',  c  :  c'; 
considering  the  rectangles  «/  as  bases  of  P  and  ^,  by  766, 

P  '.  Q  '.'.  b  :  b'. 
Therefore,  compounding  the  ratios, 

Q  \   q  '.'.  be  ',  b'<f. 


POL  YHEDRONS. 


297 


Theorem  VI. 

768.   Tivo  qtiaders  are  to  each  other  ift  the  ratio  compounded 
of  the  ratios  of  their  bases  and  altitudes. 


y\ 

7 

L 

/ 

7o 

A 

• 

7 

f 

If 

F- 

A          / 

/ 

^. 

Hypothesis.    Lei  Q  and  Q  be  two  quaders,  of  aliifude  a  and  a\ 
and  base  be  and  b'c',  respectively. 

Conclusion.     Q  :  Q  w  abc  \  a'b'cf. 

Proof.     Make  P  a  third  quader  of  altitude  a'  and  base  be. 

Then,  by  766, 

Q  :  P    '.:   a     -.a'-, 
and,  by  767, 

P  ',   q  '.'.   be   '.  b'd. 
Therefore,  compounding, 

Q  '.   Q  '.'.abc  '.  a'b'(f. 


298 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  VII. 

769.  Any  parallelopiped  is  equivalent  to  a  quader  of  equiva- 
lent base  and  equal  altitude. 


For,  supposing  AB  an  oblique  parallelopiped  on  an  oblique  base, 
prolong  the  four  edges  parallel  to  AB^  take  sect  CD  =  AB,  and  draw- 
ee J_  C£>,  and  DF  \\  CE.  Through  CE  and  /?i^pass  parallel  planes. 
Now  the  solids  DFB  and  CEA  are  congruent,  having  all  their  angles 
and  edges  respectively  equal.  Taking  each  in  turn  from  the  whole  solid 
DFAj  leaves  parallelopiped  AB  =  CD. 

In  the  line  CD  take  GH  =  CD,  and  through  G  and  H  pass  planes 
perpendicular  to  GH. 

The  solids  DFH  and  CEG  are  congruent,  therefore  parallelopiped 
CD  =  GH. 

Now  prolong  the  four  edges  not  parallel  to  GH,  and  take  LM  — 
HK,  and  through  L  and  M  pass  planes  perpendicular  to  LM. 

As  before,  parallelopiped  GH  =  LM\  but  LM  is  a  quader  of 
equivalent  base  and  equal  altitude  to  parallelopiped  AB. 


POL  YHEDRONS. 


299 


Theorem  VIII. 

770.  A  plane  passed  through  two  diagonally  opposite  edges  of 
a  parallelopiped  divides  it  into  two  equivalent  triangidar  prisms. 


fa>\ 


V 

/!x'    > 

B 

>.              ./'.'T--^^^ 

<^/Jr^^ 

JT 

5^s^-r 

n 

-k 

^ 

. '   ^ 

^ 

'•    ^« 

^*' 

2 

^^           ^    • 

7  /-"^  >^ 

K^ 

^ 

If  the  lateral  faces  are  all  rectangles,  the  two  prisms  are  congruent ; 
if  not,  the  prisms  are  still  equivalent. 

For  draw  planes  perpendicular  to  AA'  at  the  points  A  and 
A'.  Then  the  prism  ABC  A'B'C  is  equivalent  to  the  right  prism 
AEM  A'E'M',  because  the  pyramid  AEBCM'\%  congruent  to  the  pyr- 
amid A'E'B'C'M'.  In  the  same  way,  ADC  A'D'C  is  equivalent  to 
ALM  A'UM'.     But  AEM  A'E'M'  and  ALM  A'L'M'  are  congruent, 

/.    ABC  A'B'C  and  ADC  A' U C  are  equivalent. 


771.  Corollary.     Any  triangular  prism  is  half  a  parallelo- 
piped of  twice  its  base  but  equal  altitude. 


300 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem  IX. 

772.  If  a  pyramid  be  ctit  by  a  plane  parallel  to  its  base,  the 
section  is  to  the  base  as  the  square  of  tJie  perpe7idicular  on  it, 
from  the  vertex,  is  to  the  square  of  the  altitude  of  the  pyramid. 


The  section  and  base  are  similar,  since   corresponding  diagonals 
cut  them  into  triangles  similar  in  pairs  because  having  all  their  sides 
respectively  proportional. 
For 

A'C  :  AC  ::    VC  :   VC  :  :   VO"  :   VO, 
and 

B'C  :  BC  ::    VC    :    VC, 

.'.     AC    :  AC  ::  B' C  :  BC; 
and  in  the  same  way, 

B'C  :  BC  ::  A'B'  :  AB, 

.'.     t.A'B'C'^t.ABC,Q\c., 


section  :  base  :-.  A' C""  -.  AC^  w   VO^  :    VC 


773.  Corollary.  In  pyramids  of  equivalent  bases  and 
equal  altitudes,  two  sections  having  equal  perpendiculars  from 
the  vertices  are  equivalent. 


POL  YHEDRONS. 


301 


Theorem  X. 

774.     Tetrahedra    {triangular  pyramids)    having  equivalent 
bases  and  equal  altitudes  are  equivale7it. 


Divide  the  equal  altitudes  a  into  n  equal  parts,  and  through  each 
point  of  division  pass  a  plane  parallel  to  the  base. 

By  773,  all  sections  in  the  first  tetrahedron  are  triangles  equivalent 
to  the  corresponding  sections  in  the  second. 

Beginning  with  the  base  of  the  first  tetrahedron,  construct  on  each 

a 
section,  as  lower  base,  a  prism  -  high,  with  lateral  edges  parallel  to  one 

of  the  edges  of  the  tetrahedron. 

In  the  second,  similarly  construct  prisms  on  each  section,  as  upper 
base. 

Since  the  first  prism-sum  is  greater  than  the  first  tetrahedron,  and 
the  second  prism-sum  is  less  than  the  second  tetrahedron,  therefore  the 
difference  of  the  tetrahedra  is  less  than  the  difference  of  the  prism-sums. 

But,  by  771,  each  prism  in  the  second  tetrahedron  is  equivalent  to 
the  prism  next  above  it  on  the  first  tetrahedron. 

So  the  difference  of  the  prism-sums  is  simply  the  lowest  prism  of 
the  first  series.    As  n  increases,  this  decreases,  and  can  be  made  as  small 


302  THE  ELEMENTS  OF  GEOMETRY. 

as  we  please  by  taking  n  sufficiently  great ;  but  it  is  always  greater  than 
the  constant  difference  between  the  tetrahedra,  and  so  that  constant 
difference  must  be  nought. 


Theorem  XI. 

775.  A  triangular  pyramid  is  one-third  of  a  triangular  prison 
of  the  same  base  and  altitude. 


Let  E  ABC  be  a  triangular  pyramid.  Through  one  edge  of  the 
base,  as  A  C,  pass  a  plane  parallel  to  the  opposite  lateral  edge,  EB,  and 
through  the  vertex  E  pass  a  plane  parallel  to  the  base.  The  prism 
ABC  DEF  has  the  same  base  and  altitude  as  the  given  pyramid.  The 
plane  AFE  cuts  the  part  added,  into  two  triangular  pyramids,  each 
equivalent  to  the  given  pyramid;  for  E  ABC  and  A  DEF  have  the 
same  altitude  as  the  prism,  and  its  bottom  and  top  respectively  as  bases ; 
while  E  AFC  and  E  AFD  have  the  same  altitude  and  equal  bases. 


BOOK   XL 


MENSURATION,   OR   METRICAL   GEOMETRY. 
CHAPTER  I. 

THE   METRIC   SYSTEM.  —  LENGTH,   AREA. 

776.  In  practical  science,  every  quantity  is  expressed  by  a 
phrase  consisting  of  two  components,  —  one  a  number,  the 
other  the  name  of  a  thing  of  the  same  kind  as  the  quantity  to 
be  expressed,  but  agreed  on  among  men  as  a  standard  or  Unit. 

777.  The  Meastirement  of  a  magnitude  consists  in  finding 
this  number. 

778.  Measurement,  then,  is  the  process  of  ascertaining 
approximately  the  ratio  a  magnitude  bears  to  another  chosen 
as  the  standard ;  and  the  measure  of  a  magnitude  is  this  ratio 
expressed  approximately  in  numbers. 

779.  For  the  continuous-quantity  space,  the  fundamental 
unit  actually  adopted  is  the  Meter,  which  is  a  bar  of  platinum 
preserved  at  Paris,  the  bar  supposed  to  be  taken  at  the  tem- 
perature of  melting  ice. 

780.  This  material  meter  is  the  ultimate  standard  univer- 
sally chosen,  because  of  the  advantages  of  the  metric  system 

303 


304 


THE  ELEMENTS  OF  GEOMETRY. 


of  subsidiary  units  connected  with  it,  which  uses  only  decimal 
multiples  and  sub-multiples,  being  thus  in  harmony  with  the 
decimal  nature  of  the  notation  of  arithmetic. 

781.  The  metric  system  designates  multiples  by  prefixes 
derived  from  the  Greek  numerals,  and  sub-multiples  by  pi^e- 
fixes  from  the  Latin  numerals. 


Prefix. 

Meaning  as  used. 

Derivation. 

Abbreviation. 

myria- 

ten  thousand 

fJiVpLOL 

kilo- 

one  thousand 

XlXlol 

k- 

hecto- 

one  hundred 

iKarov 

h- 

deka- 

ten 

SeKa 

da- 

deci- 

one-tenth 

decem 

d- 

centi- 

one-hundredth 

centum 

c- 

milli- 

one-thousandth 

mille 

m- 

The  abbreviation  for  meter  is  m. ;  hence  km.  for  kilometer,  mm.  for  millimeter. 

782.  The  adoption  of  the  meter  gives  the  world  one  stand- 
ard sect  as  fundamental  unit. 

783.  The  Levgth  of  any  sect  is  its  ratio  to  the  meter  ex- 
pressed approximately  in  numbers. 

10 


mm 


I  decimeter  =  10  centimeters  =  100  millimeters. 


Men  of  science  often  express  their  measurements  in  terms 
of  a  subsidiary  unit,  the  Centimeter. 

The  length  of  a  sect  referred  to  the  centimeter  as  unit  is 
one  hundred  times  as  great  as  referred  to  the  meter. 


MENSURATION-,    OR  METRICAL    GEOMETRY. 


305 


784.  An  accessible  sect  may  be  practically  measured  by  the 
direct  application  of  a  known  sect,  such  as  the  edge  of  a  ruler 
suitably  divided. 

But  because  of  incommensurability,  any  description  of  a 
sect  in  terms  of  the  standard  sect  must  be  usually  imperfect 
and  merely  approximate. 

Moreover,  in  few  physical  measurements  of  any  kird  are 
more  than  six  figures  of  such  approximations  accurate ;  and 
that  degree  of  exactness  is  very  seldom  obtainable,  even  by  the 
most  delicate  instruments. 

785.  For  the  measurement  of  surfaces  the  standard  is  the 
square  on  the  linear  unit. 

786.  The  Area  of  any  surface  is  its  ratio  to  this  square. 

787.  If  the  unit  for  length  be  a  meter,  the  unit  for  area,  a 
square  on  the  meter,  is  called  a  square  meter  (m.^^ ;  better,  ^■'^). 

In  science  the  Square  Centimeter  (^'^•^)  is  adopted  as  the 
primary  unit  of  surface. 


788.    To  find  the  area  of  a  rectangle. 


3  f 


Rule.     Multiply  the  base  by  the  altitude. 

Formula.     R  =  ab. 

Proof.  —  Special  Case  :  When  the  base  and  altitude  of 
the  rectangle  are  commensurable.  In  this  case,  there  is  always 
a  sect  which  will  divide  both  base  and  altitude  exactly.  If  this 
sect  be  assumed  as  linear  unit,  the  lengths  a  and  b  are  integral 


306  THE  ELEMENTS  OF  GEOMETRY. 

numbers.  In  the  rectangle  ABCD,  divide  AD  into  ^,  and  AB 
into  b,  equal  parts.  Through  the  points  of  division  draw  lines 
parallel  to  the  sides  of  the  rectangle.  These  lines  divide  the 
rectangle  into  a  number  of  squares,  each  of  which  equals  the 
assumed  unit  of  surface.  In  the  bottom  row,  there  are  b  such 
squares ;  and,  since  there  are  a  rows,  we  have  b  squares  re- 
peated a  times,  which  gives,  in  all,  ab  squares. 

Note.  The  composition  of  ratios  includes  numerical  multiplication 
as  a  particular  case. 

But  ordinary  multiplication  is  also  an  independent  growth  from 
addition. 

In  this  latter  point  of  view,  the  multiplier  indicates  the  number  of 
additions  or  repetitions,  while  the  multiplicand  indicates  the  thing 
added  or  repeated.  This  is  not  a  mutual  operation,  and  the  product 
is  always  in  terms  of  the  unit  of  the  multiplicand.  The  multiplicand 
may  be  any  aggregate ;  the  multiplier  is  an  aggregate  of  repetitions. 
To  repeat  a  thing  does  not  change  it  in  kind,  so  the  result  is  an  aggre- 
gate of  the  same  sort  exactly  as  the  multiplicand. 

But  if  the  multiplicand  itself  is  also  an  aggregate  of  repetitions, 
the  two  factors  are  the  same  in  kind,  and  the  multiplication  is  com- 
mutative. 

This  is  the  only  sort  of  multiplication  needed  in  mensuration ;  for 
all  ratios  are  supposed  to  be  expressed  exactly  or  approximately  in  num- 
bers, and  in  our  rules  it  is  only  of  these  numbers  that  we  speak.  Thus, 
when  the  rule  says,  "  Multiply  the  base  by  the  altitude,"  it  means,  Mul- 
tiply the  number  taken  as  the  length  of  the  base,  by  the  number  which 
is  the  measure  of  the  altitude  in  terms  of  the  same  linear  unit.  The 
product  is  a  tmmber,  which  we  prove  to  be  the  area  of  the  rectangle ; 
that  is,  its  numerical  measure  in  terms  of  the  superficial  unit.  This  is 
the  meaning  to  be  assigned  whenever  in  mensuration  we  speak  of  the 
product  of  one  sect  by  another. 

General  Proof.  If  L  represent  the  unit  for  length,  and 
5  the  unit  of  surface,  and  ^  ab  the  rectangle,  the  length  of 


MENSURATION,   OR  METRICAL    GEOMETRY. 


307 


whose  base  is  b  and  the  length  of  whose  altitude  is  a^  then,  by 

542, 

C7  cib  aL     bL 

But  the  first  member  of  this  equation  is  the  area  of  the  rect- 
angle, which  number  we  may  represent  by  R ;  and  the  second 
member  is  equal  to  the  product  of  the  numbers  a  and  b ; 

.'.    R  =  ab. 


clZ 


Example  i.  Find  the  area  of  a  ribbon  I*"-  long  and  i^i"-  wdde. 

Answer,  ydit'"**  =  loC^"^-^. 

Since  a  square  is  a  rectangle  having  its  length  and  breadth 
equal,  therefore 

789.    To  find  the  area  of  a  square. 

Rule.      Take  the  second  power  of  the  mimber  denoting  the 
length  of  its  side. 

Note.    This  is  why  the  product  of  a  number  into  itself  is  called 
the  square  of  that  number. 


790.    Given,  the  area  of  a  square,  to  find  the  length  of  a  side. 

Rule.     Extract  the  square  root  of  the  number  deftoting  the 
arecL. 


308  THE  ELEMENTS  OF  GEOMETRY, 

791.  Metric  Units  of  Surface. 

I  hectar  (^^•)  =  i  sq.  hectometer  =  loooo  sq.  meters. 

I  dekar           =  =  1000  sq.  meters. 

I  ar  (a-)          =  I  sq.  dekameter  =  100  sq.  meters. 

I  deciar          =  =  10  sq.  meters. 

I  centiar         =  i  sq.  meter  =  i  sq.  meter. 

I  milliar          =  =  .1  sq.  meter. 

I  sq.  decimeter  =  .01  sq.  meter. 

I  sq.  centimeter  =  .0001  sq.  meter. 

I  sq.  millimeter  =  .000001  sq.  meter. 

Example  2.  How  many  square  centimeters  in  10  millimeters  square? 
Answer,  {iQ^'^-y  =  loo™"^-^  =  i^m.^, 

792.  Remark.  Distinguish  carefully  between  square  meters 
and  meters  square. 

We  say  10  square  kilometers  (lo^"^-^),  meaning  a  surface 
which  would  contain  10  others,  each  a  square  kilometer ;  while 
the  expression  '*  5  kilometers  square "  (s''"^-)^  means  a  square 
whose  sides  are  each  5  kilometers  long,  so  that  the  figure  con- 
tains 25^"^•^ 

Example  3.  A  square  is  looo'"-''.    Find  its  side. 


Answer,  yiooo"^-  =  31.623™- 

793.  Because  the  sum  of  the  squares  on  the  two  sides 
of  a  right-angled  triangle  is  the  square  of  the  hypothenuse, 
therefore,  also,  > 

Given,  ihe  hypothenuse  and  one  side,  to  find  the  other  side. 

Rule.  Multiply  their  sum  by  their  difference^  and  extract 
the  square  root. 

Formula,     c^  —  a^  =:i  {c  -\'  a)  {c  —  a)  =  ^^ 


MENSURATION,    OR  METRICAL    GEOMETRY.  309 

From  this  it  follows,  that,  in  an  acute-angled  triangle,  if  we 
are  given  two  sides  and  the  projection  of  one  on  the  other,  or 
two  sides  and  an  altitude,  we  can  find  the  third  side. 

Exercises.  112.  What  must  be  given  in  order  to  find  the 
medials  of  a  triangle  } 

113.  If  on  the  three  sides  of  any  triangle  squares  are 
described  outward,  the  sects  joining  their  outer  corners  are 
twice  the  medials  of  the  triangle,  and  perpendicular  to  them. 


310 


THE  ELEMENTS  OF  GEOMETRY. 


CHAPTER   II. 


RATIO   OF   ANY   CIRCLE   TO    ITS   DIAMETER. 


Problem   I. 

794.  Given^  the  perimeters  of  a  regular  inscribed  and  a  sim^ 
ilar  circiLfHscribed  polygon^  to  compute  the  perimeters  of  the 
regular  inscribed  and  circumscribed  polygons  of  double  the  tium" 
ber  of  sides. 

a        JO 


Take  AB  a  side  of  the  given  inscribed  polygon,  and  CD  a  side  of 
the  similar  circumscribed  polygon,  tangent  to  the  arc  AB  at  its  mid- 
point E. 

Join  AE,  and  at  A  and  B  draw  the  tangents  AF  and  BG;  then 
AE  is  a  side  of  the  regular  inscribed  polygon  of  double  the  number  of 
sides,  and  FG  is  a  side  of  the  circumscribed  polygon  of  double  the 
number  of  sides. 

Denote  the  perimeters  of  the  given  inscribed  and  circumscribed 
polygons  by  p  and  q  respectively,  and  the  required  perimeters  of  the 
inscribed  and  circumscribed  polygons  of  double  the  number  of  sides 
by/'  and  /  respectively. 


MENSURATION,    OR  METRICAL    GEOMETRY.  311 

Since  <9C  is  the  radius  of  the  circle  circumscribed  about  the  poly- 
gon whose  perimeter  is  q, 

.-.     q  \  p  w   OC  \   OE. 

(548.  The  perimeters  of  two  similar  regular  polygons  are  as  the  radii  of  their  cir- 
cumscribed circles.) 

But,  since  OF  bisects  the  4-  COE, 

.'.     OC  :   OE  ::   CE   :  FE, 

(523.  The  bisector  of  an  angle  of  a  triangle  divides  the  opposite  side  in  the  ratio  of 
the  other  two  sides  of  the  triangle.) 

/.     q  \  p  \\   CF  \  FE, 
whence,  by  composition, 

p  ■\-  q  '.  2p  \\   CF  '\-  FE  \  2FE  ^  CE  \  EG  \\  q  \  ^, 

since  FG  is  a  side  of  the  polygon  whose  perimeter  is  /,  and  is  con- 
tained as  many  times  in  /  as  CE  is  contained  in  q, 

.-.     p  -{-  q   :   2p   '.'.  q   '.  q'. 

If,  now,  the  letters  be  taken  to  represent  lengths  in  terms  of  the  unit 
sect  Z,  this  proportion  is 

pL  -f-  qL  :  2pL  w  qL  \  q'L, 
which  gives  the  number 

^  =  7+i-  <'> 

Again,  right  A  AEH  ~  EFN,  since  acute  ^  EAH  =  FEN, 

,\     AH  '.  AE  ::  EN  :  EF, 

.'.    p  :/    ::/  :  /, 

since  AH  and  AE  are  contained  the  same  number  of  times  in  /  and/' 
respectively,  and  EN  and  EF  are  contained  twice  that  number  of 
times  in  /'  and  /  respectively.  If,  now,  the  letters  be  taken  to  repre- 
sent length  in  terms  of  the  same  unit  sect  Z,  this  proportion  is 

pL  :  /L  : :  /Z  ;  /Z, 
which  gives  the  number  ,         , — -, 

Therefore  from  the  given  lengths  p  and  q  we  compute  q'  by  equa- 
tion (i),  and  then  with  /  and  /  we  compute  /'  by  equation  (2). 


312 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   I. 

795*     The   length   of  a  circle  whose  diameter   is   unity  is 
3.141592+. 


The  length  of  the  perimeter  of  the  circumscribed  square  is  4.     A 
side  of  the  inscribed  square  is  -|V2,  therefore  its  perimeter  is  2V 2. 

Now,  putting  p  =  2V2  and  ^  =  4  in  794,  we  find,  for  the  perime- 
ters of  the  circumscribed  and  inscribed  octagons, 

2/^ 


/  = 


=  3-3137085, 


/  =  V'//=  3.0614675. 

Then,  taking  these  as  given  quantities,  we  put  /  =  3.0614675,  and 
^  =  3-3137085,  and  find  by  the  same  formulae,  for  the  polygons  of 
sixteen  sides,  /  =  3.1825979,  and  /  =  3.1214452. 

Continuing  the  process,  the  results  will  be  found  as  in  the  following 
table :  — 


Number  of 
Sides. 

Perimeter 

of  Circumscribed 

Polygon. 

Perimeter 

of   Inscribed 

Polygon. 

4 

4.0000000 

2.8284271 

8 

3-3137085 

3.0614675 

16 
32 

3.1825979 
3-I517249 

3.1214452 
3-1365485 

64 

3.1441184 

3.1403312 

128 

3.1422236 

3-I412773 

256 

3.1417504 

3-1415138 

512 
1024 

3.I41632I 
3.1416025 

3-I415729 
3-I415877 

2048 

3-I415951 

3.1415914 

4096 

3-1415933 

3-I415923 

8192 

3.1415928 

3.1415926 

MENSURATION,    OR  METRICAL    GEOMETRY.  313 

But  since,  by  654,  a  chord  is  shorter  than  its  arc,  therefore  the 
circle  is  longer  than  any  inscribed  perimeter ;  and,  assuming  that  two 
tangents  from  an  external  point  cannot  be  together  shorter  than  the 
included  arc,  it  follows  that  the  circle  is  not  longer  than  a  circumscribed 
perimeter ;  therefore  the  circle  whose  diameter  is  unity  is  longer  than 
3.1415926  and  not  longer  than  3.1415928. 

796.  A  Variable  is  a  quantity  which  may  have  successively 
an  indefinite  number  of  different  values. 

797*  If  ^  variable  which  changes  its  value  according  to 
some  law  can  be  made  to  approach  some  fixed  value  as  nearly  as 
we  please y  but  can  never  become  equal  to  it,  the  constant  is  called 
the  Limit  of  the  variable. 

Example  4.     The  limit  of  the  fraction  -,  as  x  increases  indefitiitely, 

X 

is  zero ;  for,  by  taking  x  sufficiently  great,  we  can  make  -  less  than  any 

X 

assigned  quantity,  but  we  can  never  make  it  zero. 


PRINCIPLE   OF    LIMITS. 

798.  Ify  while  tending  toward  their  respective  limits,  two 
variables  are  always  in  the  same  ratio,  their  limits  will  have 
that  ratio. 

I 1 i hH-H M— H H^ 

^  '         y  C'  x'  C         Z'  y  Z  2 

Let  the  sects  AC  and  AL  represent  the  limits  of  any  two 
variable  magnitudes  which  are  always  in  the  same  ratio,  and 
let  Ax,  Ay,  represent  two  corresponding  values  of  the  variables 
themselves ;  then 

Ax    -.Ay    '.'.AC    :  AL. 
For  if  not,  then 

Ax  :  Ay  ::  AC  :  to  some  sect  >  or  <  AL, 


314  THE  ELEMENTS  OF  GEOMETRY, 

Suppose,  in  the  first  place,  that 

Ax   \  Ay   \\  AC   \  AL', 

where  AL'  is  less  than  AL. 

By  hypothesis,  the  variable  Ay  continually  approaches  AL, 
and  may  be  made  to  differ  from  it  by  less  than  any  given  quan- 
tity. 

Let  Ax  and  Ay,  then,  continue  to  increase,  always  remaining 
in  the  same  ratio,  until  Ay  differs  from  AL  by  less  than  the 
quantity  L'L  ;  or,  in  other  words,  until  the  point  y  passes 
the  point  Z',  and  reaches  some  point,  as  j/',  between  V  and  Z, 
and  X  reaches  the  corresponding  point  x'  on  the  sect  AC.  Then, 
since  the  ratio  of  the  two  variables  is  always  the  same, 

Ax    :  Ay    ::  Ax'    :  A/, 
But,  by  hypothesis. 

Ax    :  Ay    ::  AC   :  AL\ 

.-.     Ao^  :  A/  ::  AC    :  AL', 
But 

Ax'<AC,  .-.     Ay<AL\ 

which   is   absurd.      Hence   the   supposition   that   Ax  :  Ay  : : 
AC  :  AL'y  or  to  any  quantity  less  than  AL,  is  absurd. 
Suppose,  then,  in  the  second  place,  that 

Ax    :  Ay    ::  AC  :  AL", 

where  AL"  >  AL.     Now,  there  is  some  sect,  as  AC,  less  than 
AC,  which  is  to  AL  2iS  AC  is  to  AL". 

Substituting  this  ratio  for  that  oi  AC  to  AL'\  we  have 

Ax    '.Ay   ::  AC  :  AL, 


MENSURATION,    OR  METRICAL    GEOMETRY. 


315 


which,  by  a  process  of  reasoning  similar  to  the  above,  may  be 
shown  to  be  absurd.  Hence,  since  the  fourth  term  of  the  pro- 
portion can  be  neither  greater  nor  less  than  ALy  it  must  equal 
AL ;  that  is, 

Ax   \  Ay   w  AC   \  AL. 

799.  Corollary.     If  two  variables  are  always  equal,  their 
limits  are  equal. 

Theorem   II. 

800.  Ally  two  circles  are  to  each  other  as  their  radii. 


Proof.  By  548,  the  perimeters  of  any  two  regular  polygons  of  the 
same  number  of  sides  have  the  same  ratio  as  the  radii  of  their  circum- 
scribed circles. 

The  inscribed  regular  polygons  remaining  similar  to  each  other  when 
the  number  of  sides  is  doubled,  their  perimeters  continue  to  have  the 
same  ratio.  Assuming  the  circle  to  be  the  limit  toward  which  the 
perimeter  of  the  inscribed  polygon  increases,  by  798,  Principle  of 
Limits,  the  circles  have  the  same  ratio  as  their  radii. 


801.  Since  c  :  c' 


2r', 


c_ 
2r 


2r' 


that  is,  the  ratio  of  any  one  circle  to  its  diameter  is  the  same 
as  the  ratio  of  any  other  circle  to  its  diameter. 


3l6  THE  ELEMENTS  OF  GEOMETRY. 

This  constant  ratio  is  denoted  by  the  Greek  letter  tt.  But, 
by  795,  this  ratio  for  the  circle  with  unit  diameter,  and  there- 
fore for  every  circle,  is 

TT    =    3.I4I592-I-. 

802.   For  any  circle. 
Formula,     c  =  2r7r. 


CIRCULAR   MEASURE    OF   AN   ANGLE. 

803.  When  its  vertex  is  at  the  center  of  the  circle,  by  506, 

any  ^  _  its  intercepted  arc  _  arc 
St.  ^  semicircle  rTr' 

any  ^  _  arc 

TT 

So,  if  we  adopt  as  unit  angle  the  radian^  or  that  part  of  a  peri- 

st  ^ 
gon  denoted  by  -^^,  that  is,  the  angle  subtended  at  the  center 


Ih 


of  every  circle  by  an  arc  equal  to  its  radius,  and  hence  named 
a  radian,  then 

The  number  which  expresses  any  angle  in  radianSy  also 
expresses  its  intercepted  arc  in  terms  of  the  raditis. 

If  ti  denote  the  number  of  radians  in  any  angle,  and  /  the 
length  of  its  intercepted  arc,  then 

/ 

u  —  -, 

r 


MENSURATION,   OR  METRICAL   GEOMETRY,  317 

The  fraction  arc  divided  by  radius^  or  Uy  is  called  the  circu- 
lar measure  of  an  angle. 

804.  Arcs  are  said  to  measure  the  angles  at  the  center  which 
include  them,  because  these  arcs  contain  their  radius  as  often 
as  the  including  angle  contains  the  radian.  In  this  sense  an 
angle  at  the  center  is  measured  by  the  arc  intercepted  between 
its  sides. 


3i8 


THE  ELEMENTS  OF  GEOMETRY. 


CHAPTER   III. 


MEASUREMENT   OF    SURFACES. 


805.  By  248,  any  parallelogram  is  equivalent  to  the  rect- 
angle of   its  base  and  altitude ;   therefore, 


ziy 


To  find  the  area  of  any  parallelogram. 

Rule.     Multiply  the  base  by  the  altitude. 
Formula,     /u  =.  ab. 

806.  Corollary.     The  area  of  a  parallelogram  divided  by 
the  base  gives  the  altitude. 

807.  By  252,  any  triangle  is  equivalent  to  one-half  the  rect- 
angle of  its  base  and  altitude  ;  therefore, 


Given,  one  side  and  the  perpendicular  upon  it  from  the  opposite 
vertex,  to  find  the  area  of  a  triangle. 

Rule.     Take  half  the  product  of  the  base  into  the  altitude. 
Formula,     a  =  '^ib. 


MENSURATION,   OR  METRICAL   GEOMETRY. 


319 


808.   Given,  the  three  sides,  to  find  the  area  of  a  triangle. 


Rule.     From  half  the  sum  of  the  three  sides  subtract  each 
side  separately ;   multiply  together  the  half-sitm  and  the  three 
remainders.      The  square  root  of  this  product  is  the  area. 
Formula,     t.  :=i  \s  {s  —  a)  {s  —  b)  {s  —  c). 
Proof.     Calling  j  the  projection  of  c  on  b,  by  306, 

a'  =,  b^  j^  c"  —  2bj\ 
.      .       l^  -^  c^  -  a"" 


2b 


Calling  the  altitude  //,  this  gives 
h^  =,  c^  -  j^  =  c^  - 


{h"  -\-  c'  -  a^y 
4b^ 


/.     4h^b^  =  4^V  —  (/r*  +  ^  -  a^Yy 


2/ib  =  V^4^V2  -  (b^  ■\-  c'  -  a'Y, 


2hb 


2hb^ 

sl{2bc  -h  b^  -\-  c^  - 

-  a'){2bc  -  b' 

—  c^ 

+  «^), 

=  Vc^ 

-{-b  -\-  c){b  +  c 

-  a){a  -\-  b  — 

c){^a 

-  b  ■\- 

0, 

^  /(^  +  ^  +  0  (-^  +  ^ 

-a)(a  +  b  - 

c){a 

-b^ 

c) 

.,..  a  -\-  b  -\-  c    '        b  At  c 

Writing  s  —       \^  gives      ^ 


.-.    \hb  =  \ls{s  -  a){s  -  b){s  -  c). 
But,  by  807,  Ihb  =  A. 


320  THE  ELEMENTS  OF  GEOMETRY. 


809.    To  find  the  area  of  a  regular  polygon. 


Rule.      Take  half  the  prodtict  of  its  perimeter  by  the  radius 

of  the  inscribed  circle. 

rp 
Formula.     TV  =  — . 
2 

Proof.     Sects  from  the  center  to  the  vertices  divide  the 

polygon  into   congruent   isosceles   triangles  whose  altitude  is 

the  radius  of  the  inscribed  circle,  and  the  sum  of  whose  bases 

is  the  perimeter  of  the  polygon. 

810.    To  find  ihe  area  of  a  circle. 


Rule.     Multiply  its  squared  radius  by  -k. 
Formula.     O  ==  rV. 

If  a  regular  polygon  be  circumscribed  about  the  circle,  its 
area  TV,  by  809,  is  \rp. 


MENSURATION,   OR  METRICAL    GEOMETRY,  32 1 

If,  now,  the  number  of  sides  of  the  regular  polygon  be  con- 
tinually doubled,  the  perimeter  /  decreases  toward  c  as  limit, 
and  N  toward  circle.  But  the  variables  N  and  /  are  always  in 
the  constant  ratio  \r\  therefore,  by  798,  Principle  of  Limits, 
their  limits  are  in  the  same  ratio, 

.-.     O  =  \rc. 
But,  by  802,  c  =  2r7r, 


811.    To  find  the  area  of  a  sector. 


Rule.     Multiply  the  length  of  the  arc  by  half  the  radius. 
Formula.     5  =  J/r  =  \ur''. 
Proof.     By  506,  5 


0 

\ : 

/  : 

c  : 

\\  u  \ 

27r, 

s 

= 

0/ 
c 

= 

t 
27r 

s 

= 

c 

= 

r'^irU 
27r 

s 

= 

¥r 

= 

lur^. 

812.  An  Amnihis  is  the  figure  included  between  two  con- 
centric circles.     Its  height  is  the  difference  between  the  radii. 


322 


THE  ELEMENTS  OF  GEOMETRY. 


813.    To  find  the  area  of  a  sector  of  an  annu/us. 


Rule.  Multiply  the  sum  of  the  bounding  arcs  by  half  the 
difference  of  their  radii. 

Formula.     5  .  ^  .  =  J  (^3  —  r,)  (/,  +  4)  =  \h  (/,  +  Q. 

Proof.  The  annular  sector  is  the  difference  between  the 
two  sectors  \rj^  and  \rj^. 

But  /i  and  4  are  arcs  subtending  the  same  angle ;  therefore, 
by  804, 

^  =  4 

.-.    -1^24  -  -1^14  =  Wi  +  i^2^2  -  i^i4  —  1^,4 

=  K^. -^0(^  +  4). 
814.    To  find  the  lateral  area  of  a  prism. 


^ 


MENSURATION^   OR  METRICAL    GEOMETRY, 


323 


Rule.  Multiply  a  lateral  edge  by  the  perimeter  of  a  right 
section. 

Formula.     P  =z  Ip. 

Proof.  The  lateral  edges  of  a  prism  are  all  equal.  The 
sides  of  a  right  section,  being  perpendicular  to  the  lateral 
edges,  are  the  altitudes  of  the  parallelograms  which  form  the 
lateral  surface  of  the  prism. 

815.  A  Cylindric  Surface  is  generated  by  a  line  so  moving 
that  every  two  of  its  positions  are  parallel. 


816.  The  generatrix  in  any  position  is  called  an  Elemejtt  of 
the  surface. 

817.  A  Cylinder  is  a  solid  bounded  by  a  cylindric  surface 
and  two  parallel  planes. 


324 


THE  ELEMENTS  OF  GEOMETRY. 


8i8.  The  Axis  of  a  circular  cylinder  is  the  sect  joining  the 
centers  of  its  bases. 


.--r-^ 


819.  A  Truncated  Cylinder  is  the  portion  between  the  base 
and  a  non-parallel  section. 


820.    To  find  the  lateral  area  of  a  right  circular  cylinder. 


Rule.     Multiply  its  length  by  its  circle. 
Formula.     C  =  cl. 


MENSURATION,    OR  METRICAL    GEOMETRY. 


325 


Proof.  Imagine  the  curved  surface  slit  along  an  element 
and  then  spread  out  flat.  It  thus  becomes  a  rectangle,  having 
for  one  side  the  circle,  and  for  the  adjacent  side  an  element. 

821.  Corollary  I.  The  curved  surface  of  a  truncated 
circular  cylinder  is  the  product  of  the  circle  of  the  cylinder  by 
the  intercepted  axis. 


^vK- 


For,  by  symmetry,  substituting  an  oblique  for  the  right  sec- 
tion through  the  same  point  of  the  axis,  alters  neither  the 
curved  surface  nor  the  volume,  since  the  solid  between  the  two 
sections  will  be  the  same  above  and  below  the  right  section. 

822.  Corollary  II.  The  lateral  area  of  any  cylinder  on 
any  base  equals  the  length  of  the  cylinder  multiplied  by  the 
perimeter  of  a  right  section. 

823.  The  lateral  surface  of  a  prism  or  cylinder  is  called  its 
Mantel. 


Exercises.  114.  The  mantel  of  a  right  circular  cylinder 
is  equivalent  to  a  circle  whose  radius  is  a  mean  proportional 
between  the  altitude  of  the  cylinder  and  the  diameter  of  its 
base. 

115.  A  plane  through  two  elements  of  a  right  circular 
cylinder  cuts  its  base  in  a  chord  which  subtends  at  the  center 
an  angle  whose  circular  measure  is  u  \  find  the  ratio  of  the 
curved  surfaces  of  the  two  parts  of  the  cylinder. 


326  THE  ELEMENTS  OF  GEOMETRY. 

824.  A  Conical  Surface  is  generated  by  a  straight  line 
moving  so  as  always  to  pass  through  a  fixed  point  called  the 
apex. 


825.  A  Cone  is  a  solid  bounded  by  a  conical  surface  and  a 
plane. 


826.  A  right  circular  cone  may  be  generated  by  revolving  a 
right  triangle  about  one  pependicular.  All  the  elements  are 
equal,  and  each  is  called  the  Slant  Height  of  the  cone. 


MENSURATION,   OR  METRICAL   GEOMETRY. 


327 


827.    The  Frustum   of  a  pyramid   or  cone  is  the   portion 
included  between  the  base  and  a  parallel  section. 


828.    To  find  ihe  lateral  area  of  a  right  circular  cone. 


Rule.  Multiply  the  circle  of  its  base  by  half  the  sla7it 
height. 

Formula.     K  =  \ch  =  -n-rh. 

Proof.  If  the  curved  surface  be  slit  along  a  slant  height, 
and  spread  out  flat,  it  becomes  a  sector  of  a  circle,  with  slant 
height  as  radius,  and  the  circle  of  cone's  base  as  arc ;  therefore, 
by  811,  its  area  is  ^ch. 

829.  Corollary.  Since,  by  810,  the  base  O  =  ^cr,  there- 
fore the  curved  surface  is  to  the  base  as  the  slant  height  is 
to  the  radius  of  base,  as  the  length  of  circle  with  radius  h  is  to 
circle  with  radius  r,  as  the  perigon  is  to  the  sector  angle. 


328  THE  ELEMENTS  OF  GEOMETRY. 

830.    To  find  ihe  lateral  area  of  the  frustum  of  a  right  circular 
cone. 


Rule.  Midtiply  the  slajit  height  of  the  frustum  by  half  the 
sum  of  the  circles  of  its  bases. 

Formula.     F  ^=^  \h{c^-\'  c^  =  irh  {r^  +  r^. 

Proof.  Completing  the  cone,  and  slitting  it  along  a  slant 
height,  the  curved  surface  of  the  frustum  develops  into  the 
difference  of  two  similar  sectors  having  a  common  angle,  the 
arcs  of  the  sectors  being  the  circles  of  the  bases  of  the  frus- 
tum.    By  813,  the  area  of  this  annular  sector, 

F  =  \h(^c,  +  c). 

831.  The  axis  of  a  right  circular  cone,  or  cone  of  revolution, 
is  the  line  through  its  vertex  and  the  center  of  its  base. 

Exercises.  116.  Given  the  two  sides  of  a  right-angled 
triangle.  Find  the  area  of  the  surface  described  when  the 
triangle  revolves  about  its  hypothenuse. 

Hint.     Calling  a  and  b  the  given  altitude  and  base,  and  x  the  per- 
pendicular from  the  right  angle  to  the  hypothenuse,  by  828,  the 
area  of  the  surface  described  by  a  is  itxa,  and   by  b  is  irxb. 
Also  a  \  X  w  sla"  -^  b-"   \  b. 
117.  In  the  frustum  of  a  right  circular  cone,  on  each  base 
stands  a  cone  with  its  apex  in  the  center  of  the  other  base ; 
from  the  basal  radii  r^  and  r^^  find  the  radius  of  the  circle  in 
which  the  two  cones  cut. 


MENSURATIONy    OR  METRICAL    GEOMETRY. 


329 


Theorem   III. 

832.  The  lateral  area  of  a  friistmn  of  a  cone  of  revolution  is 
the  product  of  the  projection  of  the  frnstiims  slant  height  on  the 
axis  by  twice  tr  times  a  perpendicular  erected  at  the  mid-point  of 
this  slant  height^  and  terminated  by  the  axis. 


Proof.     By  830,  the  lateral  area  of  the  frustum  whose  slant  height 
is  PR  and  axis  MN  is 

F  =  tt{PM  -{-  RN)FR, 

But  if  Q  is  the  mid-point  of  PR,  then  PM  +  RN  =  2QO, 

.'.     P  =  27r  X  PR  X    QO. 

But  A  PRL  ~  A  QCO,  since  the  three  sides  of  one  are  drawn  per- 
pendicular to  the  sides  of  the  other ; 

.-.     PR  X   QO  =  PL  X   QC, 

/.    F  =  2it{LP  X   QC)  =  27r{MN  X   CQ). 


Exercises.  118.  Reckon  the  mantel  from  the  two  radii 
when  the  inclination  of  a  slant  height  to  one  base  is  half  a 
right  angle. 

119.  If  in  the  frustum  of  a  right  cone  the  diameter  of  the 
upper  base  equals  the  slant  height,  reckon  the  mantel  from 
the  altitude  a  and  perimeter/  of  an  axial  section. 


330 


THE  ELEMENTS  OF  GEOMETRY. 


833.    To  find  the  area  of  a  sphere. 


Rule.     Multiply  four  times  its  squared  radius  by  tt. 

Formula.     H  =  4r*7r. 

Proof.  In  a  circle  inscribe  a  regular  polygon  of  an  even 
number  of  sides.  Then  a  diameter  through  one  vertex  passes 
through  the  opposite  vertex,  halving  the  polygon  symmetrically. 

Let  PR  be  one  of  its  sides  ;  draw  PM,  RNy  perpendicular 
to  the  diameter  BD.  From  the  center  C  the  perpendicular 
CQ  bisects  PR.     Drop  the  perpendiculars  PL^  QO. 

Now,  if  the  whole  figure  revolve  about  BD  as  axis,  the 
semicircle  will  generate  a  sphere,  while  each  side  of  the  in- 
scribed polygon,  as  PR,  will  generate  the  curved  surface  of  the 
frustum  of  a  cone.     By  832,  this 

F  =  2ir{MN  X   CQ); 

and  the  sum  of  all  the  frustums,  that  is,  the  surface  of  the  solid 
generated  by  the  revolving  semi-polygon,  equals  2itCQ  into  the 
sum  of  the  projections,  BM,  MN,  NC,  etc.,  which  sum  is  BD, 

/.     Sum  of  surfaces  of  frustums  =  ^-kCQ  X  BD. 


mensuration;  or  metrical  geometry.         331 

As  we  continually  double  the  number  of  sides  of  the  inscribed 
polygon,  its  semi-perimeter  approaches  the  semicircle  as  limit, 
and  its  surface  of  revolution  approaches  the  sphere  as  limit, 
while  CQ,  its  apothem,  approaches  r,  the  radius  of  the  sphere, 
as  limit.  Representing  the  sum  of  the  surfaces  of  the  frus- 
tums by  2/s  and  BD  by  2r,  we  have 

CQ       ^ 

That  is,  the  variable  sum  is  to  the  variable  CQ  in  the  constant 
ratio  4r7r;  therefore,  by  798,  Principle  of  Limits,  their  limits 
have  the  same  ratio, 

H 

.-.     —  =  4r7r, 
r 

,'.     H  =  4r27r. 
834.  A  Calot  is  a  zone  of  only  one  base. 


835.  The  last  proof  gives  also  the  following  rule :  — 

To  find  the  area  of  a  zone. 

Rule.     Multiply  the  altitude  of  the  segment  by  twice  ir  times 
the  radius  of  the  sphere. 
Formula.     Z  =  2ar'K. 


332 


THE  ELEMENTS  OF  GEOMETRY. 


CHAPTER    IV. 

SPACE   ANGLES. 


836.  A  Plane  Angle  is  the  divergence  of  two  straight  lines 
which  meet  in  a  point. 

837.  A  Space  Angle  is  the  spread  of  two  or  more  planes 
which  meet  in  a  point. 


838.  Symmetrical  Space  Angles  are  those  which  cut  out 
symmetrical  spherical  polygons  on  a  sphere,  when  their  vertices 
are  placed  at  its  center. 


839.  A  Steregon  is  the  whole  amount  of  space  angle  round 
about  a  point  in  space. 


MENSURATION,   OR  METRICAL   GEOMETRY. 


333 


840.  A  Steradian  is  the  angle  subtended  at  the  center  by 
that  part  of  every  sphere  equal  to  the  square  of  its  radius. 

841.  The  space  angle  made  by  only  two  planes  corresponds 
to  the  lune  intercepted  on  any  sphere  whose  center  is  in  the 
common  section  of  the  two  planes. 


842.  A  Spherical  Pyramid  is  a  portion  of  a  globe  bounded 
by  a  spherical  polygon  and  the  planes  of  the  sides  of  the  poly- 
gon. The  center  of  the  sphere  is  the  apex  of  the  pyramid ; 
the  spherical  polygon  is  its  base. 


843.  Just  as  plane  angles  at  the  center  of  a  circle  are  pro- 
portional to  their  intercepted  arcs,  and  also  sectors,  so  space 


angles  at  the  center  of  a  sphere  are  proportional  to  their  inter- 
cepted  spherical  polygons y  and  also  spherical  pyramids. 


334  "^^^  ELEMENTS  OF  GEOMETRY. 

Example.  Find  the  ratio  of  the  space  angles  of  two  right  cones 
of  altitude  a^  and  a^,  but  having  the  same  slant  height,  h. 

These  space  angles  are  as  the  corresponding  calots  (or  zones  of  one 

base)  on  the  sphere  of  radius  h.    Therefore,  by  835,  the  required  ratio 

is 

2Trh{h  —  a,)  _  h  —  Uj^ 

2i:h{h  —  a^       h  —  a^ 
the  ratio  of  the  calot  altitudes. 


For  the  equilateral  and  right-angled  cones  this  becomes 

2  -  V^^ 

2  -  V2* 

844.  To  construct  a  space  angle  of  two  faces  equivalent  to 
any  polyhedral  angle,  only  involves  constructing  a  lune  equiva- 
lent to  a  spherical  polygon,  as  in  731. 

845.  To  find  the  area  of  a  lune. 

Rule.  Multiply  its  angle  in  radians  by  twice  its  squared 
radius. 

Formula.     L  =  2r^u. 

Proof.     By  703,  lunes  are  as  their  angles, 

/,     a  lune  is  to  a  hemisphere  as  its  angle  is  to  a  straight  angle, 

Z     _  u 
2r^ir        TV 

.',    L  =  2r^u, 


MENSURATION,   OR  METRICAL   GEOMETRY.  335 

846.  Corollary  I.  A  lime  measures  twice  as  many  stera- 
dians  as  its  angle  contains  radians. 

847.  Corollary  II.  If  two-faced  space  angles  are  equal, 
their  lune  angles  are  equal ;  so  a  dihedral  angle  may  be  meas- 
ured by  the  plane  angle  between  two  perpendiculars,  one  in 
each  face,  from  any  point  of  its  edge. 


848.  Suppose  the  vertex  of  a  space  angle  is  put  at  the  cen- 
ter of  a  sphere,  then  the  planes  which  form  the  space  angle  will 
cut  the  sphere  in  arcs  of  great  circles,  forming  a  spherical  poly- 
gon, whose  angles  may  be  taken  to  measure  the  dihedral  angles 
of  the  space  angle,  and  whose  sides  measure  its  face  angles. 


Hence  from  any  property  of  spherical  polygons  we  may 
infer  an  analogous  property  of  space  angles. 

For  example,  the  following  properties  of  trihedral  angles 
have  been  proved  in  our  treatment  of  spherical  triangles  :  — 

I.  Trihedral  angles  are  either  congruent  or  symmetrical 
which  have  the  following  parts  equal :  — 


336  THE  ELEMENTS  OF  GEOMETRY. 

(i)  Two  face  angles  and  the  included  dihedral  angle. 

(2)  Two  dihedral  angles  and  the  included  face  angle. 

(3)  Three  face  angles. 

(4)  Three  dihedral  angles. 

(5)  Two  dihedral  angles  and  the  face  angle  opposite  one  of 
them,  provided  the  edge  of  the  third  dihedral  angle  of  neither 
trihedral  makes  right  angles  with  any  line  in  the  half  of  the 
opposite  face  not  adjacent  to  one  of  the  face  angles  equal  by 
hypothesis. 

(6)  Two  face  angles  and  the  dihedral  angle  opposite  one  of 
them,  provided  the  other  pair  of  opposite  dihedral  angles  are 
not  supplemental. 

II.  As  one  of  the  face  angles  of  a  trihedral  angle  is  greater 
than,  equal  to,  or  less  than,  another,  the  dihedral  angle  which 
it  subtends  is  greater  than,  equal  to,  or  less  than,  the  dihedral 
angle  subtended  by  the  other. 

III.  Symmetrical  trihedral  angles  are  equivalent. 

IV.  Space  angles  having  the  same  number  and  sum  of  dihe- 
dral angles  are  equivalent. 

V.  The  face  angles  of  a  convex  space  angle  are  together 
less  than  a  perigon. 

849.  To  find  ihe  area  of  a  spherical  iriangle. 

Rule.  Multiply  its  spheHcal  excess  in  radians  by  its  squared 
radius. 

Formula,     a"  =  er^. 

Proof.  By  711,  a  triangle  is  equal  to  a  lune  whose  angle 
is  \e ;  therefore,  by  845,  'a  =  r'^e. 

850.  Corollary  I.  A  spherical  triangle  measures  as  many 
steradians  as  its  e  contains  radians. 

851.  Corollary  II.  By  714,  to  find  the  area  of  a  spherical 
polygon,  multiply  its  spherical  excess  in  radians  by  its  squared 
radius. 


MENSURATION,    OR  METRICAL    GEOMETRY. 


337 


CHAPTER  V. 

THE   MEASUREMENT    OF   VOLUMES. 

852.  The  Volume  of  a  solid  is  its  ratio  to  an  assumed  unit. 

853.  The  Unit  for  Measicremejit  of  Volume  is  a  cube  whose 
edge  is  the  unit  for  length. 


Cubic  Centimeter. 


854.  Metric  Units  for  Volume. 


Solid. 
1  kilostere 
I  hectostere 
I  dekastere 
I  stere 
I  decistere 
I  centistere 


Liquid.  Cubic. 

=  I  cubic  dekameter  = 


I  kiloliter     =  i  cubic  meter  = 

I  hectoliter  = 

I  dekaliter  = 

I  liter            =  I  cubic  decimeter    = 

I  deciliter  = 

I  centiliter  = 

I  cubic  centimeter  = 


1000  cubic  meters. 
100  cubic  meters. 
10  cubic  meters. 
I  cubic  meter. 
.1  cubic  meter. 
.01  cubic  meter. 
.001  cubic  meter. 
.0001  cubic  meter. 
.ocxx)i  cubic  meter. 
.ocxxxDi  cubic  meter. 


I  cubic  millimeter  =  .000000001  cubic  meter. 

The  authorized  abbreviation  for  cubic  is  the  index  ^  as  in  icm.3  foj.  j  cubic  centi- 
meter ;  that  for  stere  is  s.,  and  for  liter  1.. 


338 


THE  ELEMENTS  OF  GEOMETRY, 


855.  To  find  the  volume  of  a  quader. 

Rule.  Multiply  together  its  lengthy  breadth,  and  thickness. 
Proof.     By  "jG^,  two  quaclers  have  the  ratio  compounded 
of  the  ratios  of  their  bases  and  altitudes. 

856.  Corollary.  The  volume  of  any  cube  is  the  third 
power  of  the  length  of  an  edge.  This  is  why  the  third  power 
of  a  number  is  called  its  cube. 

857.  To  find  the  volume  of  any  parallelepiped . 
Rule.     Multiply  its  altitude  by  the  area  of  its  base. 
Proof.     By  769,  any  parallelopiped  is  equivalent  to  a  quader 

of  equivalent  base  and  equal  altitude. 


858.    To  find  the  volume  of  any  prism. 


Rule.     Multiply  its  altitude  by  its  base. 

Formula.      V .  P  =  aB. 

Proof.  By  771,  any  three-sided  prism  is  half  a  parallelo- 
piped, with  base  twice  the  prism's  base  and  the  same  altitude. 

Thus  the  rule  is  true  for  triangular  prisms,  and  consequently 
for  all  prisms ;  since,  by  passing  planes  through  one  lateral 
edge,  and  all  the  other  lateral  edges  excepting  the  two  adja- 
cent to  this  one,  we  can  divide  any  prism  into  triangular  prisms 
of  the  same  altitude,  whose  triangular  bases  together  make  the 
polygonal  base. 


MENSURATION,   OR  METRICAL    GEOMETRY. 


339 


859.    To  find  the  volume  of  any  cylinder. 


Rule.     Multiply  its  altitude  by  its  base. 

Proof.  The  cylinder  is  the  limit  of  an  inscribed  prism 
when  the  number  of  sides  of  the  prism  is  indefinitely  increased, 
the  base  of  the  cylinder  being  the  limit  of  the  base  of  the 
prism. 

But  always  the  variable  prism  is  to  its  variable  base  in  the 
constant  ratio  a ;  therefore,  by  798,  Principle  of  Limits,  their 
limits  will  be  to  one  another  in  the  same  ratio. 

860.    To  find  the  volume  of  any  pyramid. 


Rule.     Multiply  one-third  of  its  altitude  by  its  base. 
Formula.     Y  =  \aB. 


340  THE  ELEMENTS  OF  GEOMETRY. 

Proof.  By  775,  any  triangular  pyramid  is  one-third  of  a 
triangular  prism  of  the  same  base  and  altitude. 

The  rule  thus  proved  for  triangular  pyramids  is  true  for  all 
pyramids  ;  since  by  passing  planes  through  one  lateral  edge, 
and  all  the  other  lateral  edges  excepting  the  two  adjacent  to 
this  one,  we  can  divide  any  pyramid  into  triangular  pyramids 
of  the  same  altitude  whose  bases  together  make  the  polygonal 
base. 

861.  To  find  the  volume  of  any  cone. 


Rule.     Multiply  one-third  its  altitude  by  its  base. 

Formula  when  Base  is  a  Circle.      V.  K  =  ^ar^-n-. 

Proof.  The  base  of  a  cone  is  the  limit  of  the  base  of  an 
inscribed  pyramid,  and  the  cone  is  the  limit  of  the  pyramid. 
But  always  the  variable  pyramid  is  to  its  variable  base  in  the 
constant  ratio  \a. 

Therefore  their  limits  are  to  one  another  in  the  same  ratio, 
and  V.K  z=z  ^aB. 

Scholium.  This  applies  to  all  solids  determined  by  an 
elastic  string  stretching  from  a  fixed  point  to  a  point  describing 
any  closed  plane  figure. 


MENSURATION,    OR  METRICAL   GEOMETRY. 


341 


862.  Corollary.  The  volume  of  the  solid  generated  by 
the  revolution  of  any  triangle  about  one  of  its  sides  as  axis,  is 
one-third  the  product  of  the  triangle's  area  into  the  length  of 
the  circle  described  by  its  vertex. 


F=    fTT^A. 


PRISMATOID. 


863.  A  Prismatoid  is  a  polyhedron  whose  bases  are  any  two 
polygons  in  parallel  planes,  and  whose  lateral  faces  are  triangles 


determined  by  so  joining  the  vertices  of  these  bases  that  each 
lateral  edge,  with  the  preceding,  forms  a  triangle  with  one  side 
of  either  base. 


342 


THE  ELEMENTS  OF  GEOMETRY. 


864.  A  number  of  different  prismatoids  thus  pertain  to  the 
same  two  bases. 


865.  If  two  basal  edges  which  form  with  the  same  lateral 
edge  two  sides  of  two  adjoining  faces  are  parallel,  then  these 


two  triangular  faces  fall  in  the  same  plane,  and  together  form  a 
trapezoid. 


MENSURATION,    OR  METRICAL    GEOMETRY.  343 

866.  A  Prismoid  is  a  prismatoid  whose  bases  have  the  same 
number  of  sides,  and  every  corresponding  pair  parallel. 


867.  A  frustum  of  a  pyramid  is  a  prismoid  whose  two  bases 
are  similar. 

868.  Corollary.     Every  three-sided  prismoid  is  the  frus- 
tum of  a  pyramid. 

869.  If  both  bases  of  a  prismatoid  become  sects,  it  is  a 
tetrahedron. 


870.  A  Wedge  is  a  prismatoid  whose  lower  base  is  a  rect- 
angle, and  upper  base  a  sect  parallel  to  a  basal  edge. 


871.  The  altitude  of  a  prismatoid  is  any  sect  perpendicular 
to  both  bases. 

872.  A  Cross-Sectio7t  of  a  prismatoid  is  a  section  made  by  a 
plane  perpendicular  to  the  altitude. 


344  THE  ELEMENTS  OF  GEOMETRY. 

873.  To  find  the  volume  of  any  prismaioid. 


Rule.  Multiply  one-fourth  its  altitude  by  the  sum  of  one 
base  and  three  times  a  cross-section  at  two-thirds  the  altitude 
from  that  base. 

Formula.    D  —  -{B  -{-  z  T). 
4 

Proof.  Any  prismatoid  may  be  divided  into  tetrahedra,  all 
of  the  same  altitude  as  the  prismatoid ;  some,  as  C  FGO,  having 
their  apex  in  the  upper  base  of  the  prismatoid,  and  for  base 
a  portion  of  its  lower  base  ;  some,  as  O  ABC,  having  base  in 
the  upper,  and  apex  in  the  lower,  base  of  the  prismatoid ;  and 
the  others,  as  A  COG,  having  for  a  pair  of  opposite  edges  a 
sect  in  the  plane  of  each  base  of  the  prismatoid,  as  AC  3.nd  OG. 

Therefore,  if  the  formula  holds  good  for  tetrahedra  in  these 
three  positions,  it  holds  for  the  prismatoid,  their  sum. 

In  (i),  call  7"i  the  section  at  two-thirds  the  altitude  from 
the  base  B^ ;  then  T^  is  \a  from  the  apex.     Therefore,  by  772, 

4  4  3  3 

which,  by  860,  equals  Y,  the  volume  of  the  tetrahedron. 


MENSURATION,   OR  METRICAL    GEOMETRY. 


345 


In  (2),  B^  =  o,  being  a  point,  and  T^  is  ^a  from  the  apex ; 
/.     T,  :  B,  ::  (|«)^  :  a\  /.     T,  =  |^„ 

...     A  =  ^(^,  +  37;)  =  ^(o  +  ^^.)  =  -aB,  =  K 
4  4  3  3 


In  (3),  let  KLMN  be  the  section  T^. 


Join  CK,  CN,  OM,  ON.     Now 

A  ANK  :  A  ^(?6>  :  :  AN^   :  ^6^^   :  :  (^ay   :  a^   :  :  1   :  g. 
A  GNM:  A  (9-^C  : :   GN^   :   6^^^   :  :  (|^)^   :  tz^   :  :  4  :  9. 

But  the  whole  tetrahedron  D^  and  the  pyramid  C ANK  may 
be  considered  as  having  their  bases  in  the  same  plane,  AGO, 
and  the  same  altitude,  a  perpendicular  from  C; 

C  ANK  :  Z>3  :  :  A  ANK  :  A  AGO  :  :  i   :  9, 
.-.     C  ANK  =  4Z>3. 


346  THE  ELEMENTS  OF  GEOMETRY. 

In  same  way, 

O  GNM  \  D^   : :  A  GNM  :  A  GAC  : :  4  :  9, 

/.     OGNM=%D^. 

C  ANK  +  O  GNM    =  %D^, 

.'.     C  KLMN  +  O  KLMN  =  %D^, 
But,  by  860, 

C  KLMN  -f  O  KLMN  =  \  .  \aT,  4-  i  .  1^7;  =  \aT,, 

.-.   4_z)3  =  i^r,      /.   i^3  =  ^3r,  =  ^{B,  4-  37;), 

9  3  4  4 

since  here 

B^  =  o. 

874.  Corollary  I.     Since,  in  the  frustum  of  a  pyramid,  B 
and  T  are  similar. 


4    \  ^i'  / 


where  w^  and  ^e'^  are  corresponding  sides  of  B  and  T. 

875.  Corollary  II.     For  the  frustum  of  a  cone  of  revolu- 
tion, 


where  rj  is  the  radius  of  T. 


V,F  =  -7rr,^ 
4 


MENSURATION,    OR  METRICAL    GEOMETRY.  247 

876.  To  find  the  volume  of  a  globe. 

Rule.     Multiply  the  cube  of  its  radius  by  |-7r. 

Formula.     G  =  ^-kt^. 

Proof.  By  644,  a  tetrahedron  on  edge,  and  a  globe  with 
the  tetrahedron's  altitude  for  diameter,  have  all  their  corre- 
sponding cross-sections  equivalent  if  any  one  pair  are  equiva- 
lent. 

Hence  the  volumes  may  be  proved  equivalent,  as  in  774, 

and 

G  =  \aT, 

But  a  =  2r,  and,  by  522,  T  =  fr  .  f r .  tt, 

.-.     dP  =  I  .  2r  .  |r  .  f  r  .  TT  =  f  7rr3. 

877.  Corollary  I.  Globes  are  to  each  other  as  the  cubes 
of  their  radii. 

878.  Corollary  II.  Similar  solids  are  to  one  another  as 
the  cubes  of  any  two  corresponding  edges  or  sects. 


DIRECTION. 

879.  If  two  points  starting  from  a  state  of  coincidence  move 
along  two  equal  sects  which  do  not  coincide,  that  quality  of 
each  movement  which  makes  it  differ  from  the  other  is  its 
Direction. 

880.  If  two  equal  sects  are  terminated  at  the  same  point, 
but  do  not  coincide,  that  quality  of  each  which  makes  it  differ 
from  the  other  is  its  direction. 

881.  The  part  of  a  line  which  could  be  generated  by  a  tra- 
cing-point, starting  from  a  given  point  on  that  line,  and  moving 
on  the  line  without  ever  turning  back,  is  called  a  Ray  from 
that  given  point  as  origin. 


348  THE  ELEMENTS  OF  GEOMETRY. 

882.  Two  rays  from  the  same  origin  are  said  to  have  the 
same  direction  if  they  coincide ;  otherwise,  they  are  said  to 
have  different  directions. 

883.  Two  rays  which  have  no  point  but  the  origin  in  com- 
mon, and  fall  into  the  same  line,  are  said  to  have  opposite  direc- 
tions. 

884.  Two  rays  lying  on  parallel  lines  have  parallel-same 
directions  if  they  are  on  the  same  side  of  the  line  joining  their 
origins.     . 

885.  Two  rays  lying  on  parallel  lines  have  parallel-opposite 
directions  if  they  are  on  opposite  sides  of  the  line  joining  their 
origins. 

886.  A  sect  is  said  to  have  the  same  direction  as  the  ray  of 
which  it  is  a  part. 

887.  A  sect  is  definitely  fixed  if  we  know  its  initial  point, 
its  parallel-same  direction,  and  its  length. 

888.  The  operation  by  which  a  sect  could  be  traced  if  we 
knew  its  initial  point,  that  is,  the  operation  of  carrying  a 
tracing-point  in  a  certain  parallel-same  direction  until  it 
passes  over  a  given  number  of  units  for  length,  is  called  a 
Vector. 

889.  The  position  of  B  relative  to  A  is  indicated  by  the 
length  and  parallel-same  direction  of  the  sect  AB  drawn  from 
A  to  B.  If  you  start  from  A,  and  travel,  in  the  direction  indi- 
cated by  the  ray  from  A  through  B^  and  traverse  the  given 
number  of  units,  you  get  to  B.  This  parallel-same  direction 
and  length  may  be  indicated  equally  well  by  any  .other  sect,  such 
as  A' B\  which  is  equal  to  AB  and  in  parallel-same  direction. 

890.  As  indicating  an  operation,  the  vector  ^^  is  completely 
defined  by  the  parallel-same  direction  and  length  of  the  trans- 
ferrence.  All  vectors  which  are  of  the  same  magnitude  and 
parallel-same  direction,  and  only  those,  are  regarded  as  equal. 

Thus  AB  is  not  equal  to  BA. 


MENSURATION^   OR  METRICAL    GEOMETRY.  349 


PRINCIPLE    OF    DUALITY. 

JOIN    OF    POINTS    AND    OF   LINES. 

891.  The  line  joining  two  points  is  called  the  _/(?/;/  of  the 
Two  Points.  The  point  common  to  two  intersecting  lines  is 
called  the  Join  of  the  Two  Lines. 

892.  Pencil  of  Lines.  A  fixed  point,  A^  may  be  joined  to 
all  other  points  in  space. 

We  get  thus  all  the  lines  which  can  be  drawn  through  the 
point  A.  The  aggregate  of  all  these  lines  is  called  a  Pencil  of 
Lines.  The  fixed  point  is  called  the  Base  of  the  pencil.  Any 
one  of  these  lines  is  said  to  be  a  line  in  the  pencil,  and  also  to 
be  a  line  in  the  fixed  point.  In  this  sense,  we  say  not  only 
that  a  point  may  lie  in  a  line,  but  also  that  a  line  may  lie  in  a 
point,  meaning  that  the  line  passes  through  the  point. 

893.  In  most  cases,  we  can,  when  one  figure  is  given,  con- 
struct another  such  that  lines  take  the  place  of  points  in  the 
first,  and  points  the  place  of  lines. 

Any  theorem  concerning  the  first  thus  gives  rise  to  a  corre- 
sponding theorem  concerning  the  second  figure.  Figures  or 
theorems  related  in  this  manner  are  called  Reciprocal  Figures 
or  Recipi'ocal  Theorems. 

894.  Small  letters  denote  lines,  and  the  join  of  two  elements 
is  denoted  by  writing  the  letters  indicating  the  elements,  to- 
gether. 

Thus  the  join  of  the  points  A  and  B  is  the  line  AB,  while 
ab  denotes  the  join,  or  point  of  intersection,  of  the  lines  a  and  b. 


ROW   OF    points,    pencil    OF    LINES. 

895.  A  line  contains  an  infinite  number  of  points,  called  a 
Row  of  Points,  of  which  the  line  is  the  Base, 
A  row  is  all  points  in  a  line. 


350 


THE  ELEMENTS  OF  GEOMETRY. 


The  reciprocal  figure  is  all  lines  in  a  pointy  or  all  lines  pass- 
ing through  the  point. 

A  Flat  Pe7icil  is  the  aggregate  of  all  lines  in  a  plane  which 
pass  through  a  given  point.  In  plane  geometry,  by  a  pencil  we 
mean  a  flat  pencil. 

RECIPROCAL   THEOREMS. 


8961.  A  point  moving  along  a 
line  describes  a  row. 

8971.  A  sect  is  a  part  of  a  row 
described  by  a  point  moving  from 
one  position,  A^  to  another  posi- 
tion, B. 


896'.  A  line  turning  about  a 
point  describes  a  pencil. 

897'.  An  angle  is  a  part  of  a 
pencil  described  by  a  line  turning 
from  one  position,  a,  to  another 
position,  b. 


Thus  to  sect  AB  corresponds  4-  ^^• 

LINKAGE. 
898.  The  Peancellier  Cell  consists  of  a  rhombus  movably 


jointed,  and  two  equal  links  movably  pivoted  at  a  fixed  point, 
and  at  two  opposite  extremities  of  the  rhombus. 


MENSURATION,    OR  METRICAL    GEOMETRY. 


351 


TO    DRAW   A    STRAIGHT    LINE. 

899.  Take  an  extra  link,  and,  while  one  extremity  is  on  the 
fixed  point  of  the  cell,  pivot  the  other  extremity  to  a  fixed 
point.  Then  pivot  the  first  end  to  one  of  the  free  angles  of 
the  rhombus.  The  opposite  vertex  of  the  rhombus  will  now 
describe  a  straight  line,  however  the  linkage  be  pushed  or 
moved. 

Proof.  By  the  bar  FD,  the  point  D  is  constrained  to  move 
on  the  circle  ADR  ;  therefore  -^  ADR^  being  the  angle  in  a 
semicircle,  is  always  right. 


If,  now,  E  moves  on  EM  ±  AMy 

A  ADR  ~  A  AME, 
.-.     DA  :  AR    :  :  AM  :  AE, 
/.    DA  ,  AE  =  RA  ,  AM. 


Therefore,  if  AE  .  AD  is  constant,  E  moves  on  the  straight 
line  EAf.     But  because  BDCE  is  a  rhombus,  and  AB  =:  AC, 

.*.     D  and  iV  are  always  on  the  variable  sect  AE. 


352  THE  ELEMENTS  OF  GEOMETRY. 

Always 

AB^  =  AN^  +  NB%         and        ~BE'  =  EN^  +  NB\ 

:.    AB^  -  ~BE'  =■  AN'  -  JV£^ 

=  {AN  +  iV^^)  (^iV^  -  NE)  =  ^^  .  AD. 

900.  Corollary.  The  efficacy  of  our  cell  depends  on  its 
power  to  keep,  however  it  be  deformed,  the  product  of  two 
varying  sects  a  constant. 


Therefore  a  Hart  four-bar  cell,  constructed  as  in  the  accom- 
panying figure,  may  be  substituted  for  the  Peancellier  six-bar 
cell,  since  AE  .  AD  equals  a  constant. 

Also,  for  the  Hart  cell  may  be  substituted  the  quadruplane, 
four  pivoted  planes. 

CROSS-RATIO. 

901.  If  four  points  are  collinear,  two  may  be  taken  as  the 
extremities  of  a  sect,  which  each  of  the  others  divides  inter- 
nally or  externally  in  some  ratio. 

The  ratio  of  these  two  ratios  is  called  the  Cross-Ratio  of 
the  four  points. 

The  cross-ratio  ——  :  -— —  is  written  {ABCD). 
CB     DB 

Distinguishing  the  "step"  AB  from  BA,  as    of   opposite 

**  sense,"  and  taking  the  points  in  the  two  groups  of  two  in  a 

definite  order,  to  write  out  a  cross-ratio,  make  first   the   two 


MENSURATION,    OR   METRICAL    GEOMETRY.  353 

bars,  and  put  crossing  these  the  letters  of  the  first  group  of 

A       A 
two,  thus  — -  :  — - ;  then  fill  up,  crosswise,  the  first  by  the  first 
B        B 

letter  of  the  second  group  of  two,  the  second  by  the  second. 

If  we  take  the  points  in  a  different  order,  the  value  of  the 
cross-ratio  may  change. 

We  can  do  this  in  twenty-four  different  ways  by  forming  all 
permutations  of  the  letters. 

But  of  these  twenty-four  cross-ratios,  groups  of  four  are 
equal,  so  that  there  are  really  only  six  different  ones. 

We  have  the  following  rules  :  — 

I.  If  in  a  cross-ratio  the  two  groups  be  interchanged^  its  value 

remains  unaltered. 

{ABCD)  =  {CDAB). 

II.  If  in  a  cross-ratio  the  tzvo  points  belonging  to  one  of  the 
two  groups  be  interchanged^  the  cross-ratio  cha?iges  to  its  recip- 
rocal. 

I 


(ABCD) 


{ABDC) 


1.  and  II.  are  proved  by  writing  out  their  values. 

III.  Fjvm  II.  it  follozuSj  that,  if  we  inteixhange  the  elements 
in  each  pair^  the  cross-ratio  remains  unaltered. 

{ABCD)  =  {BADC). 

IV.  If  in  a  cross-ratio  the  two  middle  letters  be  interchanged^ 
the  cross-ratio  changes  into  its  complement. 

{ABCD)  =  I  -  {ACBD), 

This  is  proved  by  taking  the  step-equation  for  any  four  col- 
linear  points, 

BC  .AD  ^  CA  .  BD  +  AB  .  CD  =  o, 

and  dividing  it  by  CB  .  AD. 


354 


THE  ELEMENTS  OF  GEOMETRY. 


Theorem   IV. 

go2.  If  any  four  concurrent  lines  are  cut  by  a  transversal, 
any  cross-ratio  of  the  four  points  of  intersection  is  constant  for 
all  positions  of  the  transversal. 


Hypothesis.  Let  A,  B,  C,  D,  be  the  intersection  points  for  f, 
the  transversa/  in  one  posit/on,  and  let  A\  B',  C ,  jy,  be  the  cor- 
responding intersection  points  for  t\  the  transversal  in  another 
position. 

Conclusion.     {ABCD)  =  {A'B'Cjy);  that  is, 

AC  ,  AD  ^  A[C_  .  A'D' 
CB  '  DB       C'B'  '  jyB'' 

Proof.  Through  the  points  A  and  B  on  t  draw  parallels  to  /', 
which  cut  the  concurrent  lines  in  C2,  D2,  B^,  and  A„  C^,  D^. 

AACC^-^  £^BCC„ 


AC    ^   AC^ 
CB  C,B' 


and 
and 


A  ADD:,  ~  A  BDD,, 


AD 
DB 


AD, 
D,B' 


where  account  is  taken  of  "  sense.' 


MENSURATION,    OR  METRICAL    GEOMETRY.  355 

Hence 

CB  *  Z>B       C,B    '   D,B       AD^    '    Z>,b'' 

but 

ii^  =  f^'  and  C^^C:^^ 

AD,      A'ly  D,B      jyB" 

.      AC  ,  AD  ^  A'C  .  CB'  ^  y^^C^  .  A'jy 
"      CB'  DB      AD  '  DB'       CB'  ''  D'B'' 


EXERCISES. 


BOOK    I. 


1 20.  Show  how  to  make  a  rhombus  having  one  of  its  diagonals 
equal  to  a  given  sect. 

121.  If  two  quadrilaterals  have  three  consecutive  sides,  and  the  two 
contained  angles  in  the  one  respectively  equal  to  three  consecutive  sides 
and  the  two  contained  angles  in  the  other,  the  quadrilaterals  are  congru- 
ent. 

122.  Two  circles  cannot  cut  one  another  at  two  points  on  the  same 
side  of  the  line  joining  their  centers. 

123.  Prove,  by  an  equilateral  triangle,  that,  if  a  right-angled  triangle 
have  one  of  the  acute  angles  double  of  the  other,  the  hypothenuse  is 
double  of  the  side  opposite  the  least  angle. 

1 24.  Draw  a  perpendicular  to  a  sect  at  one  extremity. 

125.  Draw  three  figures  to  show  that  an  exterior  angle  of  a  triangle 
may  be  greater  than,  equal  to,  or  less  than,  the  interior  adjacent  angle. 

126.  Any  two  exterior  angles  of  a  triangle  are  together  greater  than 
a  straight  angle. 

127.  The  perpendicular  from  any  vertex  of  an  acute-angled  triangle 
on  the  opposite  side  falls  within  the  triangle. 

128.  The  perpendicular  from  either  of  the  acute  angles  of  an  obtuse- 
angled  triangle  on  the  opposite  side  falls  outside  the  triangle. 

129.  The  semi-perimeter  of  a  triangle  is  greater  than  any  one  side^ 
and  less  than  any  two  sides. 

3S7 


358  THE  ELEMENTS  OF  GEOMETRY. 

130.  The  perimeter  of  a  quadrilateral  is  greater  than  the  sum,  and 
less  than  twice  the  sum,  of  the  diagonals. 

131.  If  a  triangle  and  a  quadrilateral  stand  on  the  same  base,  and 
the  one  figure  fall  within  the  other,  that  which  has  the  greater  surface 
shall  have  the  greater  perimeter. 

132.  If  one  angle  of  a  triangle  be  equal  to  the  sum  of  the  other  two, 
the  triangle  can  be  divided  into  two  isosceles  triangles. 

133.  If  any  sect  joining  two  parallels  be  bisected,  this  point  will 
bisect  any  other  sect  drawn  through  it  and  terminated  by  the  parallels. 

134.  Through  a  given  point  draw  a  line  such  that  the  part  inter- 
cepted between  two  given  parallels  may  equal  a  given  sect. 

135.  The  medial  from  vertex  to  base  of  a  triangle  bisects  the  inter- 
cept on  every  parallel  to  the  base. 

136.  Show  that  the  surface  of  a  quadrilateral  equals  the  surface  of  a 
triangle  which  has  two  of  its  sides  equal  to  the  diagonals  of  the  quadri- 
lateral, and  the  included  angle  equal  to  either  of  the  angles  at  which  the 
diagonals  intersect. 

137.  Describe  a  square,  having  given  a  diagonal. 

138.  ABC  is  a  right-angled  triangle;  BCED  is  the  square  on  the 
hypothenuse;  ACKH  and  ABFG  are  the  squares  on  the  other  sides. 
Find  the  center  of  the  square  ABFG  (which  may  be  done  by  drawing 
the  two  diagonals),  and  through  it  draw  two  lines,  one  parallel  to  BC, 
and  the  other  perpendicular  to  BC.  This  divides  the  square  ABFG 
into  four  congruent  quadrilaterals.  Through  each  mid-point  of  the  sides 
of  the  square  BCED  draw  a  parallel  to  AB  or  A C.  If  each  be  extended 
until  it  meets  the  second  of  the  other  pair,  they  will  cut  the  square 
BCED  into  a  square  and  four  quadrilaterals  congruent  to  ACKH  and 
the  four  quadrilaterals  in  ABFG. 

139.  The  orthocenter,  the  centroid,  and  the  circumcenter  of  a  tri- 
angle are  collinear,  and  the  sect  between  the  first  two  is  double  of  the 
sect  between  the  last  two. 

140.  The  perpendicular  from  the  circumcenter  to  any  side  of  a 
triangle  is  half  the  sect  from  the  opposite  vertex  to  the  ortho- 
center. 

141.  Sects  drawn  from  a  given  point  to  a  given  circle  are  bisected ; 
find  the  locus  of  their  mid-points. 


EXERCISES.  359 


142.  The  intersection  of  the  Hnes  joining  the  mid-points  of  opposite 
sides  of  a  quadrilateral  is  the  mid-point  of  the  sect  joining  the  mid-points 
of  the  diagonals. 

143.  A  parallelogram  has  central  symmetry. 


SYMMETRY. 

144.  No  triangle  can  have  a  center  of  symmetry,  and  every  axis  of 

symmetry  is  a  medial. 

145.  Of  two  sides  of  a  triangle,  that  is  the  greater  which  is  cut  by 
the  perpendicular  bisector  of  the  third  side. 

146.  If  a  right-angled  triangle  is  symmetrical,  the  axis  bisects  the 
right  angle. 

147.  An  angle  in  a  triangle  ^s-ill  be  acute,  right,  or  obtuse,  according 
as  the  medial  through  its  vertex  is  greater  than,  equal  to,  or  less  than, 
half  the  opposite  side. 

148.  If  a  quadrilateral  has  axial  symmetry,  the  number  of  vertices 
not  on  the  axis  must  be  even ;  if  none,  it  is  a  symmetrical  trapezoid ;  if 
two,  it  is  a  kite. 

149.  A  kite  has  the  following  seven  properties ;  from  each  prove  all 
the  others  by  proving  that  a  quadrilateral  possessing  it  is  a  kite. 

(i)  One  diagonal,  the  axis,  is  the  perpendicular  bisector  of  tlie 
other,  which  will  be  called  the  transverse  axis. 

(2)  The  axis  bisects  the  angles  at  the  vertices  which  it  joins. 

(3)  The  angles  at  the  end-points  of  the  transverse  axis  are  equal,  and 
equally  divided  by  the  latter. 

(4)  Adjacent  sides  which  meet  on  the  axis  are  equal. 

(5)  The  axis  divides  the  kite  into  two  triangles  which  are  congruent, 
with  equal  sides  adjacent. 

(6)  The  transverse  axis  divides  the  kite  into  two  triangles,  each  of 
which  is  symmetrical. 

(7)  The  Hnes  joining  the  mid-points  of  opposite  sides  meet  on  the 
axis,  and  are  equally  inclined  to  it. 

150.  A  symmetrical  trapezoid  has  the  following  five  properties ;  from 
each  prove  all  the  others  by  proving  that  a  quadrilateral  possessing  it  is 
a  symmetrical  trapezoid. 


360  THE  ELEMENTS  OF  GEOMETRY. 

(i)  Two  opposite  sides  are  parallel,  and  have  a  common  perpendicu- 
lar bisector. 

(2)  The  other  two  opposite  sides  are  equal,  and  equally  inclined  to 
either  of  the  other  sides. 

(3)  Each  angle  is  equal  to  one,  and  supplemental  to  the  other,  of 
its  two  adjacent  angles. 

(4)  The  diagonals  are  equal,  and  divide  each  other  equally. 

(5)  One  median  line  bisects  the  angle  between  those  sides  produced 
which  it  does  not  bisect,  and  likewise  bisects  the  angle  between  the  two 
diagonals. 

151.  Prove  the  properties  of  the  parallelogram  from  its  central  sym- 
metry. 

152.  A  kite  with  a  center  is  a  rhombus;  a  symmetrical  trapezoid 
with  a  center  is  a  rectangle ;  if  both  a  rhombus  and  a  rectangle,  it  is  a 
square. 


BOOK    II. 

153.  The  perpendicular  from  the  centroid  to  a  line  outside  the  tri- 
angle equals  one-third  the  sum  of  the  perpendiculars  to  that  line  from 
the  vertices. 

154.  If  two  sects  be  each  divided  internally  into  any  number  of 
parts,  the  rectangle  contained  by  the  two  sects  is  equivalent  to  the  sum 
of  the  rectangles  contained  by  all  the  parts  of  the  one,  taken  separately, 
with  all  the  parts  of  the  other. 

155.  The  square  on  the  sum  of  two  sects  is  equivalent  to  the  sum 
of  the  two  rectangles  contained  by  the  sum  and  each  of  the  sects. 

156.  The  square  on  any  sect  is  equivalent  to  four  times  the  square 
on  half  the  sect. 

157.  The  rectangle  contained  by  two  internal  segments  of  a  sect 
grows  less  as  the  point  of  section  moves  from  the  mid-point. 

158.  The  sum  of  the  squares  on  the  two  segments  of  a  sect  is  least 
when  they  are  equal. 

159.  If  the  hypothenuse  of  an  isosceles  right-angled  triangle  be 
divided  into  internal  or  external  segments,  the  sum  of  their  squares  is 


EXERCISES.  361 


equivalent  to  twice  the  square  on  the  sect  joining  the  point  of  section  to 
the  right  angle. 

160.  Describe  a  rectangle  equivalent  to  a  given  square,  and  having 
one  of  its  sides  equal  to  a  given  sect. 

161.  Find  the  locus  of  the  vertices  of  all  triangles  on  the  same  base, 
having  the  sum  of  the  squares  of  their  sides  constant. 

162.  The  center  of  a  fixed  circle  is  the  point  of  intersection  of  the 
diagonals  of  a  parallelogram ;  prove  that  the  sum  of  the  squares  on 
the  sects  drawn  from  any  point  on  the  circle  to  the  four  vertices  of  the 
parallelogram  is  constant. 

163.  Thrice  the  sum  of  the  squares  on  the  sides  of  any  pentagon  is 
equivalent  to  the  sum  of  the  squares  on  the  diagonals,  together  with  four 
times  the  sum  of  the  squares  on  the  five  sects  joining,  in  order,  the  mid- 
points of  those  diagonals. 

164.  The  sum  of  the  squares  on  the  sides  of  a  triangle  is  less  than 
twice  the  sum  of  the  rectangles  contained  by  every  two  of  the  sides. 

165.  If  from  the  hypothenuse  of  a  right-angled  triangle  sects  be 
cut  off  equal  to  the  adjacent  sides,  the  square  of  the  middle  sect  thus 
formed  is  equivalent  to  twice  the  rectangle  contained  by  the  extreme 
sects. 

BOOK    III. 

166.  From  a  point,  two  equal  sects  are  drawn  to  a  circle.  Prove 
that  the  bisector  of  their  angle  contains  the  center  of  the  circle. 

167.  Describe  a  circle  of  given  radius  to  pass  through  a  given  point 
and  have  its  center  in  a  given  line. 

168.  Equal  chords  in  a  circle  are  all  tangent  to  a  concentric  circle. 

169.  Two  concentric  circles  intercept  equal  sects  on  any  common 
secant. 

1 70.  If  two  equal  chords  intersect  either  within  or  without  a  circle, 
the  segments  of  the  one  equal  the  segments  of  the  other. 

171.  Divide  a  circle  into  two  segments  such  that  the  angle  in  the 
one  shall  be  seven  times  the  angle  in  the  other. 

172.  ABC  and  ABC  zx^  two  triangles  such  that  AC  =  AC  ', 
prove  the  circle  through  A,  B,  C,  equal  to  that  through  A,  B,  C, 


362  THE  ELEMENTS   OF  GEOMETRY. 

1 73.  Pass  a  circle  through  four  given  points.  When  is  the  problem 
possible  ? 

174.  \i  AC  and  BD  be  two  equal  arcs  in  a  circle  ABCD,  prove 
chord  AB  parallel  to  chord  CD. 

175.  Circles  described  on  any  two  sides  of  a  triangle  as  diameters 
intersect  on  the  third  side,  or  the  third  side  produced. 

1 76.  If  one  circle  be  described  on  the  radius  of  another  as  diameter, 
any  chord  of  the  larger  from  the  point  of  contact  is  bisected  by  the 
smaller. 

177.  If  two  circles  cut,  and  from  one  of  the  points  of  intersection 
two  diameters  be  drawn,  their  extremities  and  the  other  point  of  inter- 
section will  be  collinear. 

178.  Find  a  point  inside  a  triangle  at  which  the  three  sides  shall 
subtend  equal  angles. 

179.  On  the  produced  altitudes  of  a  triangle  the  sects  between  the 
orthocenter  and  the  circumscribed  circle  are  bisected  by  the  sides  of 
the  triangle. 

180.  If  on  the  three  sides  of  any  triangle  equilateral  triangles  be 
described  outwardly,  the  sects  joining  their  circumcenters  form  an  equi- 
lateral triangle. 

181.  If  two  chords  intersect  at  right  angles,  the  sum  of  the  squares 
on  their  four  segments  is  equivalent  to  the  square  on  the  diameter. 

182.  The  opposite  sides  of  an  inscribed  quadrilateral  are  produced 
to  meet ;  prove  that  the  bisectors  of  the  two  angles  thus  formed  are  at 
right  angles. 

183.  The  feet  of  the  perpendiculars  drawn  from  any  point  in  its 
circumscribed  circle  to  the  sides  of  a  triangle  are  collinear. 

184.  From  a  point  P  outside  a  circle,  two  secants,  PAB,  PDC,  are 
drawn  to  the  circle  ABCD ;  AC,  BD,  are  joined,  and  intersect  at  O. 
Prove  that  O  lies  on  the  chord  of  contact  of  the  tangents  drawn  from  P 
to  the  circle.  Hence  devise  a  method  of  drawing  tangents  to  a  circle 
from  an  external  point  by  means  of  a  ruler  only. 

185.  A  variable  chord  of  a  given  circle  passes  through  a  fixed  point ; 
find  the  locus  of  its  mid-point. 

186.  Find  the  locus  of  points  from  which  tangents  to  a  given  circle 
contain  a  given  angle. 


EXERCISES.  363 

187.  The  hypothenuse  of  a  right-angled  triangle  is  given;  find  the 
loci  of  the  comers  of  the  squares  described  outwardly  on  the  sides  of 
the  triangle. 

BOOK  IV. 

188.  If  a  quadrilateral  be  circumscribed  about  a  circle,  the  sum  of 
two  opposite  sides  is  equal  to  the  sum  of  the  other  two. 

189.  Find  the  center  of  a  circle  which  shall  cut  off  equal  chords 
from  the  three  sides  of  a  triangle. 

190.  Draw  a  line  which  would  bisect  the  angle  between  two  hnes 
which  are  not  parallel,  but  which  cannot  be  produced  to  meet. 

191.  The  angle  between  the  altitude  of  a  triangle  and  the  line 
through  vertex  and  circumcenter  equals  half  the  difference  of  the  angles 
at  the  base. 

192.  The  bisector  of  the  angle  at  the  vertex  of  a  triangle  also  bisects 
the  angle  between  the  altitude  and  the  line  through  vertex  and  circum- 
center. 

193.  If  an  angle  bisector  contains  the  circumcenter,  the  triangle  is 
isosceles. 

194.  If  a  circle  can  be  inscribed  in  a  rectangle,  it  must  be  a  square. 

195.  If  a  rhombus  can  be  inscribed  in  a  circle,  it  must  be  a  square. 

196.  The  square  on  the  side  of  a  regular  pentagon  is  greater  than 
the  square  on  the  side  of  the  regular  decagon  inscribed  in  the  same 
circle  by  the  square  on  the  radius. 

197.  The  intersections  of  the  diagonals  of  a  regular  pentagon  are 
the  vertices  of  another  regular  pentagon ;  so  also  the  intersections  of 
the  alternate  sides. 

198.  The  sum  of  the  five  angles  formed  by  producing  the  alternate 
sides  of  a  regular  pentagon  equals  a  straight  angle. 

199.  The  circle  through  the  mid-points  of  the  sides  of  a  triangle, 
called  the  medioscribed  circle,  passes  also  through  the  feet  of  the  alti- 
tudes, and  bisects  the  sects  between  the  orthocenter  and  vertices. 

Hint.  Let  ABC  be  the  triangle ;  H^  A',  Z,  the  mid-points  of  its  sides ;  Xy 
V,  Z,  the  feet  of  its  altitudes ;  O  its  orthocenter ;  C/,  F,  JV,  the  mid-points 
of  AO,  BO,  CO. 

LHWU  is  a  rectangle ;  so  is  HKUV\  and  the  angles  at  X^  Y,  Z,  are  right. 


364  THE  ELEMENTS  OF  GEOMETRY. 

200.  The  mediocenter  is  midway  between  the  circumcenter  and  the 
orthocenter. 

201.  The  diameter  of  the  circle  inscribed  in  aright-angled  triangle, 
together  with  the  hypothenuse,  equals  the  sum  of  the  other  two  sides. 

202.  An  equilateral  polygon  circumscribed  about  a  circle  "is  equi- 
angular if  the  number  of  sides  be  odd. 

203.  Given  the  vertical  angle  of  a  triangle  and  the  sum  of  the 
sides  containing  it ;  find  the  locus  of  the  circumcenter. 

204.  Given  the  vertical  angle  and  base  of  a  triangle;  find  the 
locus  of 

(i)  The  center  of  the  inscribed  circle. 

(2)  The  centers  of  the  escribed  circles. 

(3)  The  orthocenter. 

(4)  The  centroid. 

(5)  The  mediocenter. 

205.  Of  all  rectangles  inscribable  in  a  circle,  show  that  a  square  is 
the  greatest. 

BOOK  VI. 

206.  The  four  extremities  of  two  sects  are  concyclic  if  the  sects  cut 
so  that  the  rectangle  contained  by  the  segments  of  one  equals  the  rect- 
angle contained  by  the  segments  of  the  other. 

207.  If  two  circles  intersect,  and  through  any  point  in  their  common 
chord  two  other  chords  be  drawn,  one  in  each  circle,  their  four  extrem- 
ities are  concyclic. 

208.  Does  the  magnitude  of  the  third  proportional  to  two  given 
sects  depend  on  the  order  in  which  the  sects  are  taken  ? 

209.  Through  a  given  point  inside  a  circle  draw  a  chord  so  that  it 
shall  be  divided  at  the  point  in  a  given  ratio. 

210.  If  two  triangles  have  two  angles  supplemental  and  other  two 
equal,  the  sides  about  their  third  angles  are  proportional. 

211.  Find  two  sects  from  any  two  of  the  six  following  data:  their 
sum,  their  difference,  the  sum  of  their  squares,  the  difference  of  their 
squares,  their  rectangle,  their  ratio. 


EXERCISES.  365 


212.  If  two  sides  of  a  triangle  be  cut  proportionally,  the  lines 
drawn  from  the  points  of  section  to  the  opposite  vertices  will  intersect 
on  the  medial  from  the  third  vertex. 

213.  Given  the  base  of  a  triangle  and  the  ratio  of  the  two  sides; 
find  the  locus  of  the  vertex. 


MISCELLANEOUS. 

214.  Find  the  locus  of  a  point  at  which  two  adjacent  sides  of  a 
rectangle  subtend  supplementary  angles. 

215.  Draw  through  a  given  point  a  line  making  equal  angles  with 
two  given  Hnes. 

216.  From  a  given  point  place  three  given  sects  so  that  their 
extremities  may  be  in  the  same  line,  and  intercept  equal  sects  on  that 
line. 

217.  Divide  a  sect  into  two  parts  such  that  the  square  of  one  of  the 
parts  shall  be  half  the  square  on  the  whole  sect. 

218.  Find  the  locus  of  the  point  at  which  a  given  sect  subtends  a 
given  angle. 

219.  Given  a  curve,  to  ascertain  whether  it  is  an  arc  of  a  circle  or 
not. 

220.  If  two  opposite  sides  of  a  parallelogram  be  bisected,  and  lines 
be  drawn  from  these  two  points  of  bisection  to  the  opposite  angles, 
these  lines  will  be  parallel,  two  and  two,  and  will  trisect  both  diagonals. 

221.  Through  two  given  points  on  opposite  sides  of  a  line  draw 
lines  to  meet  in  it  silch  that  the  angle  they  form  is  bisected. 

222.  The  squares  on  the  diagonals  of  a  quadrilateral  are  double 
of  the  squares  on  the  sides  of  the  parallelogram  formed  by  joining  the 
mid-points  of  its  sides. 

223.  If  a  quadrilateral  be  described  about  a  circle,  the  angles  sub- 
tended at  the  center  by  two  opposite  sides  are  supplemental. 

224.  Two  circles  touch  at  A  and  have  a  common  tangent  BC. 
Show  that  BA  C  is  a  right  angle. 

225.  Find  the  locus  of  the  point  of  intersection  of  bisectors  of  the 
angles  at  the  base  of  triangles  on  the  same  base,  and  having  a  given 
vertical  angle. 


366  THE  ELEMENTS   OF  GEOMETRY. 

226.  Find  the  locus  of  the  centers  of  circles  which  touch  a  given 
circle  at  a  given  point. 

227.  Two  equal  given  circles  touch  each  other,  and  each  touches 
one  side  of  a  right  angle ;  find  the  locus  of  their  point  of  contact. 

228.  In  a  given  Hne  find  a  point  at  which  a  given  sect  subtends  a 
given  angle. 

229.  Describe  a  circle  of  given  radius  to  touch  a  given  line  and 
have  its  center  on  another  given  line. 

230.  At  any  point  in  the  circle  circumscribing  a  square,  show  that 
one  of  the  sides  subtends  an  angle  thrice  the  others. 

231.  Divide  a  given  arc  of  a  circle  into  two  parts  which  have  their 
chords  in  a  given  ratio. 

232.  The  sect  of  a  common  tangent  between  its  points  of  contact 
is  a  mean  proportional  between  the  diameters  of  two  tangent  circles. 

233.  Any  regular  polygon  inscribed  in  a  circle  is  a  mean  propor- 
tional between  the  inscribed  and  circumscribed  regular  polygons  of  half 
the  number  of  sides. 

234.  The  circumcenter,  the  centroid,  the  mediocenter,  and  the 
orthocenter  form  a  harmonic  range. 


EXERCISES   IN   GEOMETRY  OF  THREE  DIMENSIONS  AND 
IN   MENSURATION. 

The  most  instructive  problems  in  geometry  of  three  dimensions  are 
made  by  generalizing  those  first  solved  for  plane  geometry.  This  way 
of  getting  a  theorem  in  solid  geometry  is  often  difficult,  but  a  number 
of  the  exercises  here  given  are  specially  adapted  for  it. 

In  the  author's  "Mensuration"  (published  by  Ginn  &  Co.)  are  given 
one  hundred  and  six  examples  in  metrical  geometry  worked  out  com- 
pletely, and  five  hundred  and  twenty-four  exercises  and  problems,  of 
which  also  more  than  twenty  are  solved  completely,  and  many  others 
have  hints  appended. 


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